KC Sinha Mathematics Solution Class 11 Chapter 8 Sine and Cosine के अनुप्रयोग (Application of Sine and Cosine) Exercise 8.1

Exercise 8.1

Question 1

Sol :
∠A+∠B+∠C=180°

2x+3x+7x=180°

12x=180°

$x=\frac{180^{\circ}}{12}$

∠A=2x=2×15=30°

∠B=3x=3×15=45°

∠C=7x=7×15=105°

sine formula:

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

a=ksinA , b=ksinB , c=ksinC

a=ksin30° , b=ksin45° , c=ksin105°

$a=\frac{k}{2}$ , $b=\frac{k}{\sqrt{2}}$ , c=ksin(90°+15°)

c=kcos15°

c=kcos(45°-30°)

c=k(cos45°cos30°+sin45°sin30°)

$c=k\left(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}\right)$

$c=k \frac{(\sqrt{3}+1)}{2 \sqrt{2}}$


$\frac{a}{b}=\frac{\frac{k}{2}}{\frac{k}{\sqrt{2}}}$

$=\frac{\sqrt{2}}{2}$


$\frac{b}{c}=\frac{\frac{k}{\sqrt{2}}}{k \frac{(\sqrt{3}+1)}{2 \sqrt{2}}}$

$=\frac{2}{\sqrt{3}+1}$


∴a:b:c=√2:2:(√3+1)

Question 2

Sol :
$\frac{a}{\sin A}=\frac{b}{\sin B}$

$\frac{4}{\sin A}=\frac{12}{\sin 30^{\circ}}$

$\frac{4}{\sin A}=\frac{12}{\frac{1}{2}}$

$\frac{4}{\sin 2 x}=24$

$\frac{4}{24}=\sin A$

$\frac{1}{6}=\sin A$

Question 3

Sol :
Sine formula:
$\frac{a}{\sin A}=\frac{b}{\sin B}$

$\frac{a}{\sin [180^{\circ}-(B+C)]}=\frac{b}{\sin B}$

$\frac{a}{\sin (B+C)}=\frac{b}{\sin B}$

$\frac{\sin B}{\sin (B+C)}=\frac{b}{a}$

Question 4

Sol :
∠A+∠B+∠C=180°

30°+45°+∠C=180°

75°+∠C=180°

∠C=105°

sine formula:
$\frac{a}{\sin a}=\frac{b}{\sin b}=\frac{c}{\sin c}$

$\frac{a}{\sin 30^{\circ}}=\frac{b}{\sin 45^{\circ}}=\frac{\sqrt{3}+1}{\sin 105^{\circ}}$

$\frac{a}{\frac{1}{2}}=\frac{b}{\frac{1}{\sqrt{2}}}=\frac{\sqrt{3}+1}{\frac{\sqrt{3}+1}{2 \sqrt{2}}}$

2a=b√2=2√2


CASE-1
2a=2√2

a=√2

CASE-II
b√2=2√2

b=2

Question 5

Sol :
∠A+∠B=60°..(i) ∠C=30°

∠A+∠B+∠C=180°

∠A+∠B+30°=180°

∠A+∠B=180°-30°

∠A+∠B=150°..(ii)

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∠A-∠B=60°

$\begin{aligned} &\angle A-\angle B=60^{\circ}\\ &\begin{array}{r} \angle A+\angle B=150^{\circ} \\ \hline 2 \angle A=210 \end{array} \end{aligned}$

∠A=105° , ∠B=45°

sine formula:
$\frac{a}{\sin A}=\frac{b}{\sin B}$

$\frac{a}{b}=\frac{\sin A}{\sin B}$

$\frac{a}{b}=\frac{\sin 105^{\circ}}{\sin 45^{\circ}}$

$\frac{a}{b}=\frac{\frac{\sqrt{3}+1}{2 \sqrt{2}}}{\frac{1}{\sqrt{2}}}$

a:b=(√3+1):2

Question 6

Sol :
∠A=7x , ∠B=2x , ∠C=x

∠A+∠B+∠C=180°

7x+2x+x=180°

10x=180°

x=18°

∠A=7x=7×18°=126°
 ∠C=x=18°

sine formula:
$\frac{a}{\sin A}=\frac{c}{\sin C}$

$\frac{\sin c}{\sin A}=\frac{c}{a}$

$\frac{\sin 18^{\circ}}{\sin 126^{\circ}}=\frac{c}{a}$

$\frac{\sin 18^{\circ}}{\sin (90^{\circ}+36^{\circ})}=\frac{c}{a}$

$\frac{\sin 18^{\circ}}{\cos 36^{\circ}}=\frac{c}{a}$

[∵$\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$

 $\cos 36^{\circ}=\frac{\sqrt{5}+1}{4}$]

$\frac{\frac{\sqrt{5}-1}{4}}{\frac{\sqrt{5+1}}{4}}=\frac{c}{a}$

$\frac{\sqrt{5}-1}{\sqrt{5}+1} \times \frac{\sqrt{5}-1}{\sqrt{5}-1}=\frac{c}{a}$

$\frac{5-\sqrt{5}-\sqrt{5}+1}{(\sqrt{5})^{2}-1^{2}}=\frac{c}{a}$

$\frac{6-2 \sqrt{3}}{5-1}=\frac{c}{a}$

$\frac{2(3-\sqrt{5})}{4}=\frac{c}{a}$

(3-√5):2=c:a

Question 7

Sol :
$\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$

$\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}$

$\cos B=\frac{2^{2}+(\sqrt{3}-1)^{2}-(\sqrt{6})^{2}}{2 \times 2 \times(\sqrt{3}-1)}$

$\cos B=\frac{4+(\sqrt{3})^{2}+1^{2}-2 \sqrt{3} \cdot 1-6}{4(\sqrt3-1)}$

$\cos B=\frac{4+3+1-2 \sqrt{3}-6}{4(\sqrt{3}-1)}$

$\cos B=\frac{-2 \sqrt{3}+2}{4(\sqrt{3}-1)}$

$=\frac{-2(\sqrt{3}-1)}{4(\sqrt{3}-1)}=\frac{1}{2}$

cosB=-cos60°

cosB=cos(180°-60°)

cosB=cos120°

B=120°

Question 8

Sol :
cosine formula:
$\cos B=\frac{c^{2}+a^{2}-b^{2}}{2 c a}$

$\cos 60^{\circ}=\frac{c^{2}+a^{2}-b^{2}}{2 c a}$

$\frac{1}{2}=\frac{c^{2}+a^{2}-b^{2}}{2c a}$

c²+a²-b²=ca

c²+a²-ca=b2

c²+a²-2ca+ca=b²

(c-a)²=b²-ca

Question 9

Sol :
L.H.S

$\frac{b^{2}+c^{2}-a^{2}}{c^{2}+a^{2}-b^{2}}$

[]

$=\frac{\frac{b^{2}+c^{2}-a^{2}}{2 a b c}}{\frac{c^{2}+a^{2}-b^{2}}{2 a b c}}$

$=\frac{\frac{b^{2}+c^{2}-a^{2}}{2 b c \times a}}{\frac{c^{2}+a^{2}-b^{2}}{2 \operatorname{ca}\times b}}$

$=\frac{\frac{\cos A}{a}}{\frac{\cos B}{b}}$

$=\frac{\cos A}{a} \times \frac{b}{\cos B}$

$\because \left[\frac{a}{\sin A}=\frac{b}{\sin B}\right]$

$=\frac{\cos A}{\frac{b \sin A}{\sin B}} \times \frac{b}{\cos B}$

$=\frac{\cos A}{\sin A} \times \frac{\sin B}{\cos B}$

=cotA×tanB

$=\frac{\tan B}{\tan A}$

Question 10

Sol :
a=3 , b=5 , c=7

$\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$

$\cos C=\frac{3^{2}+5^{2}-7^{2}}{2 \times 3 \times 5}$

$\cos C=\frac{9+25-49}{30}$

$\cos C=\frac{-15}{30}$

cos C=-cos60°

cosC=cos(180°-60°)

cosC=cos(120°)

C=120°

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Question 11

Sol :

a=8 , b=10 , c=12

$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$

$=\frac{10^{2}+12^{2}-8^{2}}{2 \times 10 \times 12}$

$=\frac{100+144-64}{240}$

$=\frac{180^{\circ}}{240^{\circ}}=\frac{3}{4}$


$\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$

$=\frac{8^{2}+10^{2}-12^{2}}{2 \times 8 \times 10}$

$=\frac{64+100-144}{160}=\frac{20^{\circ}}{160^{\circ}}$

$2 \cos ^{2} \frac{c}{2}-1=\frac{1}{8}$

$2 \cos ^{2} \frac{c}{2}=\frac{1}{8}+1$

$2 \cos ^{2} \frac{c}{2}=\frac{1+8}{8}$

$\cos ^{2} \frac{c}{2}=\frac{9}{16}$

$\cos \frac{c}{2}=\pm \sqrt{\frac{9}{16}}$

[$\because \cos \frac{c}{2}>0$]

$\cos \frac{c}{2}=\frac{3}{4}$

$\therefore \cos A=\cos \frac{c}{2}$

$A=\dfrac{c}{2}$

⇒2A=C

Question 12

Sol :
$c^{4}-2\left(a^{2}+b^{2}\right) c^{2}+a^{4}+a^{2} b^{2}+b^{4}=0$

$c^{4}-2\left(a^{2}+b^{2}\right) c^{2}+\left(a^{2}\right)^{2}+2 a^{2} \cdot b^{2}+\left(b^{2}\right)^{2}-a^{2} b^{2}=0$

$\left(c^{2}\right)^{2}-2\left(a^{2}+b^{2}\right) c^{2}+\left(a^{2}+b^{2}\right)^{2}=a^{2} b^{2}$

$\left(a^{2}+b^{2}-c^{2}\right)^{2}=a^{2} b^{2}$

$\left(a^{2}+b^{2}-c^{2}\right)^{2}=\frac{1}{4} \times 4 a^{2} b^{2}$

$\frac{\left(a^{2}+b^{2}-c^{2}\right)^{2}}{4 a^{2} b^{2}}=\frac{1}{4}$

$\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}=\frac{1}{4}$

$\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\pm \sqrt{\frac{1}{4}}$

$\cos C=\pm \frac{1}{2}$

CASE-I

$\cos C=\frac{1}{2}$

cosC=cos60°

C=60°


CASE-II

$\cos C=-\frac{1}{2}$

cos C=-cos60°

cos C=cos(180°-60°)

cos C=cos(120°)

C=120°

Question 13

Sol :

(a+b+c)(b+c-a)=3bc

(b+c+a)(b+c-a)=3bc

$(b+c)^{2}-a^{2}=3 b c$

$b^{2}+c^{2}+2 b c-a^{2}=3 b c$

$b^{2}+c^{2}-a^{2}=b c$

$b^{2}+c^{2}-a^{2}=\frac{1}{2} \times 2 b c$

$\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{1}{2}$

cos A=cos60°

A=60°

Question 14

Sol :
$a^{4}+b^{4}+c^{4}=2 c^{2}\left(a^{2}+b^{2}\right)$

$a^{4}+b^{4}+c^{4}=2 c^{2} a^{2}+2 b^{2} c^{2}$

$a^{4}+b^{4}+c^{4}-2 b^{2} c^{2}-2 c^{2} a^{2}=0$

[]

$a^{4}+b^{4}+c^{4}+2 a^{2} b^{2}-2 b^{2} c^{2}-2 c^{2} a^{2}=2 a^{2} b^{2}$

$\left(a^{2}\right)^{2}+\left(b^{2}\right)^{2}+\left(c^{2}\right)^{2}+2 a^{2} b^{2}-2 b^{2} c^{2}-2 c^{2} a^{2}=2 a^{2} b^{2}$

$\left(a^{2}+b^{2}-c^{2}\right)^{2}=2 a^{2} b^{2}$

$\left(a^{2}+b^{2}-c^{2}\right)^{2}=\frac{1}{2} \times {4} a^{2} b^{2}$

$\frac{\left(a^{2}+b^{2}-c^{2}\right)^{2}}{4 a^{2} b^{2}}=\frac{1}{2}$

$\left[\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right]^{2}=\frac{1}{2}$

$\cos ^{2} C =\frac{1}{2}$

$\cos C=\pm \sqrt{\frac{1}{2}}$


$\cos C=\pm \frac{1}{\sqrt{2}}$

CASE-I
$\cos C=\frac{1}{\sqrt{2}}$

cos C=cos 45°


c=45°


CASE-II

$\cos C=-\frac{1}{\sqrt{2}}$

cos C=-cos45°

cos C=cos(180°-45°)

cos C=cos 135°

C=135°

Question 15

Sol :
$\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k$

b+c=11k , c+a=12k , a+b=13k

2a+2b+2c=11k+12k+13k

2(a+b+c)=36k

a+b+c=18k

a=7k , b=6k , c=5k

$\frac{\cos A}{7}=\frac{c^{2}+b^{2}-a^{2}}{2 b c \cdot 7}$

$=\frac{(5 k)^{2}+(6 k)^{2}-(7 k)^{2}}{2 \times 6 k \times 5 k \times 7}$

$=\frac{25 k^{2}+36 k^{2}-49 k^{2}}{2 \times 6 k \times 5 k \times 7}$

$=\frac{12 k^{2}}{12 k^{2} \times 35}=\frac{1}{35}$

$\frac{\cos B}{19}=\frac{c^{2}+a^{2}-b^{2}}{2ca\times 19}$

$=\frac{(5 k)^{2}+(7 k)^{2}-(6 k)^{2}}{2 \times 5 k \times 7 k \times 19}$

$=\frac{25 k^{2}+49 k^{2}-36 k^{2}}{38 k^{2} \times 35}$

$=\frac{38 k^{2}}{38 k^{2}+35}=\frac{1}{35}$

$\frac{\cos C}{25}=\frac{a^{2}+b^{2}-c^{2}}{2 a b \times 25}$

$=\frac{(7 k)^{2}+(6 k)^{2}-(5 k)^{2}}{2 \times 7 k \times 6 k \times 25}$

$=\frac{49 k^{2}+36 k^{2}-25 k^{2}}{2 \times 7 k \times 6 k \times 25}$

$=\frac{60 k^{2}}{2 \times 7 \times 6 \times k^{2} \times 25}=\frac{1}{35}$

$\therefore \frac{\cos A}{7}=\frac{cos B}{19}=\frac{cos C}{25}$

Question 16

Sol :
∵A+B+C=180°

2B+B=180°

3B=180°

$B=\dfrac{180^{\circ}}{3}=60^{\circ}$

$\cos B=\frac{c^{2}+a^{2}-b^{2}}{2 c a}$

$\cos 60^{\circ}=\frac{c^{2}+a^{2}-b^{2}}{2 c a}$

$\frac{1}{2}=\frac{c^{2}+a^{2}-b^{2}}{2 c a}$

$c^{2}+a^{2}-b^{2}=c a$

$a^{2}-c a+c^{2}=b^{2}$

R.H.S

$\frac{a+c}{\sqrt{a^{2}-a c+c^{2}}}=\frac{a+c}{\sqrt{b^{2}}}$

$=\frac{a+c}{b}$

[∵ we know that ]

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

a=ksinA , b=ksinB , c=ksinC

Now , $\frac{a+c}{b}=\frac{k \sin A+k \sin }{k\sin B}$

$=\frac{k( \sin A+\sin C)}{K \sin B}$

$\frac{a+c}{\sqrt{a^{2}-a c+c^{2}}}=\frac{2 \sin\frac{A+C}{2} \cos \frac{A-C}{2}}{\sin B}$

$=\frac{2 \sin \left(\frac{180^{\circ}-B}{2}\right) \cos \frac{A-C}{2}}{\sin B}$

$\frac{a+c}{\sqrt{a^{2}-a c+c^{2}}}=\frac{2 \sin \left(\frac{180^{\circ}}{2}-\frac{B}{2}\right) \cos \frac{A-C}{2}}{\sin B}$

$=\frac{2 \cos \frac{B}{2} \cos \frac{A-C}{2}}{2+\frac{B}{2} \cos \frac{B}{2}}$

$\frac{a+c}{\sqrt{a^{2}-a c+c^{2}}}=\frac{\cos \frac{A-C}{2}}{\sin \frac{60^{\circ}}{2}}$

$\frac{a+c}{\sqrt{a^{2}-a c+c^{2}}}=\frac{\cos \frac{A-C}{2}}{\frac{1}{2}}$

$\frac{a+c}{\sqrt{a^{2}-a c+c^{2}}}=2 \cos \frac{A-C}{2}$

Question 17

Sol :

a+b=√3(a-b)

$\frac{1}{\sqrt{3}}=\frac{a-b}{a+b}$

$\tan \frac{A-B}{2}=\frac{a-b}{a+b} \cdot \cot \frac{c}{2}$

$\tan \frac{A-B}{2}=\frac{1}{\sqrt{3}} \times \cot \frac{60^{\circ}}{2}$

$\tan \frac{A-B}{2}=\frac{1}{\sqrt3} \times \sqrt{3}$

$\frac{\tan A-B}{2}=\tan 45^{\circ}$

$\frac{A-B}{2}=45^{\circ}$

⇒A-B=90°

Question 18

Sol :
(i)
L.H.S
$2\left(a \sin ^{2} \frac{C}{2}+\operatorname{csin}^{2} \frac{A}{2}\right)=c+a-b$

$=2\left[a \times \frac{(s-a)(s-b)}{a b}+c \frac{(s-b)(s-c)}{b c}\right]$

$=2\left[\frac{(s-a)(s-b)+(s-b)(s-c)}{b}\right]$

$=2\left[\frac{(s-b)[s-a+s-c)}{b}\right]$

$=2\left[\frac{(s-b)(2 s-a-c)}{b}\right]$

$=2\left[\frac{(s-b)\left(a+b+c-a-c\right)}{b}\right]$

$=\frac{2(s-b) \cdot b}{b}$

=a+b+c-2b

=a+c-b


(ii) $b \cos ^{2} \frac{C}{2}+c \cos ^{2} \frac{B}{2}=\frac{1}{2}(a+b+c)$
Sol :
L.H.S
$b \cos ^{2} \frac{C}{2}+c \cos ^{2} \frac{B}{2}$

$={b \cdot \frac{s(s-c)}{a b}}+c \cdot \frac{s(s-b)}{a c}$

$=\frac{s(s-c)+s(s-b)}{a}$

$=\frac{S(s-c+s-b)}{a}$

$=\frac{s(2s-c-b)}{a}$

$=\frac{S\left(a+b+ c-c-b\right)}{a}$

$=\frac{S \times a}{a}$

$=\frac{1}{2}(a+b+c)$


(iii) $\frac{\cos A}{\operatorname{ccos} B+b \cos C}+\frac{\cos B}{a \cos C+\cos A}+\frac{\cos C}{b \cos A+a \cos B}=\frac{a^{2}+b^{2}+c^{2}}{2 a b c}$

Sol :
L.H.S
$\frac{\cos A}{\operatorname{ccos} B+b \cos C}+\frac{\cos B}{a \cos C+\cos A}+\frac{\cos C}{b \cos A+a \cos B}$

$=\frac{\frac{b^{2}+c^{2}-a^{2}}{2 b c}}{\frac{a}{1}}+\frac{\frac{a^{2}+c^{2}-b^{2}}{2 a c}}{\frac{b}{1}}+\frac{\frac{a^{2}+b^{2}-c^{2}}{2 a b}}{\frac{c}{1}}$

$=\frac{b^{2}+c^{2}-a^{2}}{2 a b c}+\frac{a^{2}+c^{2}-b^{2}}{2 a b c}+\frac{a^{2}+b^{2}-c^{2}}{2 a b c}$

$=\frac{b^{2}+c^{2}-a^{2}+a^{2}+c^{2}-b^{2}+a^{2}+b^{2}-c^{2}}{2 a b c}$

$=\frac{a^{2}+b^{2}+c^{2}}{2 a b c}$

Question 19

(i) $a(\cos B+\cos C-1)+b(\cos C+\cos A-1)+c(\cos A+\cos B-1)=0$
Sol :
L.H.S
=$a(\cos B+\cos C-1)+b(\cos C+\cos A-1)+c(\cos A+\cos B-1)$

=acosB+acosC-a+bcosc+bcosA-b+ccosA+ccosB-c

=(acosB+bcosA)+(acosC+ccosA)+(bcosC+ccosB)-a-b-c

=c++b+a-a-b-c

=0

(ii) acos(A+B+C)-bcos(B+A)-ccos(A+C)=0
Sol :
L.H.S
=acos(A+B+C)-bcos(B+A)-ccos(A+C)

=acos180°-bcos(180°-C)-ccos(180°-B)

=a(-1)-b(-cosC)-c(-cosB)

=a(-1)-bcosC+ccosB

=-a+a

=0

Question 20

(i) $\left(b^{2}-c^{2}\right) \cot A+\left(c^{2}-a^{2}\right) \cot B+\left(a^{2}-b^{2}\right) \cot C=0$
Sol :

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

a=ksinA , b=ksinB , c=ksinC

$\left(b^{2}-c^{2}\right) cot A=\left(k^{2} \sin^{2} B+k^2 \sin ^{2} C\right)\cot A$

$=k^{2}\left(\sin ^{2} B-\sin ^{2} C\right) \cot A$

$=k^{2} \sin (B+c) \sin (B-c) \frac{\cos A}{\sin A}$

$=k^{2} \sin (B+c) \sin (B-C) \cdot \sin(B-C).\dfrac{\cos A}{\sin(180^{\circ}-(B+C))}$

$=k^{2} \sin (B+C) \sin (B-C) \frac{\cos A}{\sin (B+C)}$

$=k^{2}[\sin B \cos C-\cos B \sin C] \cos A$

$\left(b^{2}-c^{2}\right) \cot A=k^{2}\left[\cos A \sin B \cos C-\cos A \cos B \sin C\right]$

$\left(c^{2}-a^{2}\right) \cot B=k^{2} \sin (C-A) \cos B$

$=k^{2}(\sin C \cos A-\cos C \sin A) \cos B$

$\left(c^{2}-a^{2}\right) \cot B=k^{2}\left(\cos A \cos B \sin C-\sin A \cos B \cos C\right)$..(ii)

$\left(a^{2}-b^{2}\right) cot C=k^{2} \sin (A-B) \cos C$

$=k^{2}[\sin A \cos B-\cos A\sin B] \cos c$

$\left(a^{2}-b^{2}\right)\cot=k^{2}\left[\sin A \cos B \cos C -\cos A \sin B \cos C\right]$..(iii)

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$\left(b^{2}-c^{2}\right)\cot A+\left(c^{2}-a^{2}\right) \cot B+\left(a^{2}-b^{2}\right) \cot C$

$=k^{2} \times 0$

=0


(iii) $\frac{1+\cos (A-B) \cos C}{1+\cos (A-C) \cos B}=\frac{a^{2}+b^{2}}{a^{2}+c^{2}}$
Sol :
$=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

a=ksinA , b=ksinB  , c=ksinC

L.H.S
$\frac{1+\cos (A-B) \cos C}{1+\cos (A-C) \cos B}$

$=\frac{1+\cos (A-B) \cos \left(180^{\circ}-(A+B)\right.}{1+\cos (A-C) \cos \left[180^{\circ}-(A+C)\right]}$

$=\frac{1+\cos (A-B)\left[-\cos (A+B)\right]}{1+\cos(A-C)[-\cos (A+C)]}$

$=\frac{1-\cos (A-B) \cos (A+B)}{1-\cos(A-C) \cos (A+C)}$

$=\frac{1-\left(\cos ^{2} A-\sin ^{2} B\right)}{1-\left(\cos ^{2} A-\sin ^{2} C\right)}$

$=\frac{1-\cos^{2} A+\sin ^{2} B}{1-\cos ^{2} A+\sin ^{2} C}$

$=\frac{\sin ^{2} \theta+\sin ^{2} B}{\sin^{2} A+\sin ^{2} C}$

$=\frac{\frac{a^{2}}{k^{2}}+\frac{b^{2}}{k^{2}}}{\frac{a^{2}}{k^{2}}+\frac{c^{2}}{k^{2}}}$

$=\frac{\frac{a^{2}+b^{3}}{k^{2}}}{\frac{a^{2}+c^{2}}{k^{2}}}$

$=\frac{a^{2}+b^{2}}{a^{2}+c^{2}}$


(iv) $\frac{b^{2}-c^{2}}{\cos B+\cos C}+\frac{c^{2}-a^{2}}{\cos C+\cos A}+\frac{a^{2}-b^{2}}{\cos A+\cos B}=0$
Sol :
$=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

a=ksinA , b=ksinB  , c=ksinC

$\frac{b^{2}-c^{2}}{\cos B+\cos c}=\frac{k^{2} \sin ^{2} B-k^{2} \sin ^{2} C}{\cos B+\cos C}$

$=\frac{k^{2}\left[\sin ^{2} B-\sin ^2 C\right]}{\cos B+\cos C}$

$=k^{2} \frac{\left[\left(1-\cos ^{2} B\right)-\left(1-\cos ^{2}C\right)\right.}{\cos B+\cos c}$

$=\frac{k^{2}\left[1-\cos ^{2} B-1+\cos ^{2} C\right]}{\cos B+\cos C}$

$=\frac{k^{2}\left[\cos ^{2} C-\cos ^{2} B\right]}{\cos B+\cos C}$

$=\frac{k^{2}(\cos C+\cos B)(\cos C-\cos B)}{\cos B+\cos C}$

$\frac{b^{2}-c^{2}}{\cos B+\cos C}=k^{2}(\cos C-\cos B)$..(i)

$\frac{c^{2}-a^{2}}{\cos C+\cos A}=k^{2}(\cos A-\cos C)$..(ii)

$\frac{a^{2}-b^{2}}{\cos A+ \cos B}=k^{2}[\cos B-\cos A]$..(iii)

$\frac{b^{2}-c^{2}}{\cos B+\cos C}+\frac{c^{2}-a^{2}}{\cos C+\cos A}+\frac{a^{2}-b^{2}}{\cos A+\cos B}=k^{2} \times 0=0$


(v) $\frac{a^{2} \sin (B-C)}{\sin B+\sin C}+\frac{b^{2} \sin (C-A)}{\sin C+\sin A}+\frac{c^{2} \sin (A-B)}{\sin A+\sin B}=0$
Sol :

$=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

a=ksinA , b=ksinB  , c=ksinC

$\frac{a^{2} \sin (B-C)}{\sin B+\sin C}=\frac{k^{2} \sin ^{2} A \sin (B-C)}{\sin B+\sin C}$

$=\frac{k^{2} \sin A \sin A \sin (B-C)}{\sin B+\sin C}$

$=\frac{k^{2} \sin A \sin [180^{\circ}-(B+C)] \sin (B \cdot C)}{\sin B+\sin C}$

$=\frac{k^{2} \sin A \sin (B+C) \sin (B-C)}{\sin B+\sin C}$

$=\frac{k^{2} \sin A\left(\sin ^{2} \theta-\sin ^{2} c\right)}{\sin B+\sin C}$

$=\frac{k^{2} \sin A(\sin B+\sin C)(\sin B-\sin C)}{\sin B+\sin C}$

$\frac{a^{2} \sin (B-C)}{\sin B+\sin C}=k^{2} \sin A(\sin B-\sin C)$..(i)

$\frac{b^{2} \sin (C-A)}{\sin C+\sin A}=k^{2} \sin B(\sin C-\sin A)$..(ii)

$\frac{c^{2} \sin (A-B)}{\sin A+\sin B}=k^{2} \sin C(\sin A-\sin B)$..(iii)

$\frac{a^{2} \sin (B-C)}{\sin B+\sin C}+\frac{b^{2} \sin (C-A)}{\sin C+\sin A}+\frac{c^{2} \sin (A-B)}{\sin A+\sin B}$

$=k^{2}[\sin A \sin B-\sin A \sin C+\sin B \sin C-\sin A +\sin A \sin C-\sin B \sin C]$

$=k^{2} \times 0=0$


(vi) $\frac{b^{2}-c^{2}}{a^{2}}=\frac{\sin (B-C)}{\sin (B+C)}$
Sol :
$=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

a=ksinA , b=ksinB  , c=ksinC

L.H.S
$\frac{b^{2}-c^{2}}{a^{2}}=\frac{k^{2}\sin^{2} B^{2}-k^{2} \sin ^{2} C}{k^{2} \sin ^{2} A}$

$=\frac{k^{2}\left(\sin ^{2} B-\sin ^{2} c\right)}{k^{2} \sin ^{2}[180^{\circ}-(B+c)]}$

$=\frac{\sin (B+C) \sin (B-C)}{\sin ^{2}(B+C)}$

$=\frac{\sin (B-C)}{\sin (B+C)}$


(vii) $\tan \left(\frac{\mathrm{A}}{2}+\mathrm{B}\right)=\frac{c+b}{c-b} \tan \frac{\mathrm{A}}{2}$
Sol :
R.H.S
$\frac{c+b}{c-b} \tan \frac{A}{2}=\frac{1}{\frac{c-b}{c+b} \cot \frac{A}{2}}$

$=\frac{1}{\tan \frac{C-B}{2}}$

$=\cot \left(\frac{C-B}{2}\right)$

$=\cot \left(\frac{C}{2}-\frac{B}{2}\right)$

[∵ A+B+C=180°
C=180°-A-B
=180°-(A+B)]

$=\cot \left(\frac{180^{\circ}-A-B}{2}-\frac{B}{2}\right)$

$=\cot \left(\frac{180^{\circ}}{2}-\frac{A}{2}-\frac{B}{2}-\frac{B}{2}\right)$

$=\cot\left(90^{\circ}-\frac{A}{2}-B\right)$

$=\cot \left[90^{\circ}-\left(\frac{A}{2}+B\right)\right]$

$=\tan \left(\frac{A}{2}+B\right)$

(viii)
Sol :
$=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

a=ksinA , b=ksinB  , c=ksinC

$b+c=2 a \cos \frac{B-C}{2}$

$\frac{b+c}{2 a}=\cos \frac{B-C}{2}$

$\frac{k \sin B+k \sin C}{2 k \sin A}=\cos \frac{B-C}{2}$

$\frac{k(\sin B+\sin C)}{2 k \sin A}=\cos \frac{B-C}{2}$

$\frac{2 \sin \frac{B+ C}{2} \cos \frac{B-C}{2}}{2 \sin A}=\cos \frac{B-C}{2}$

$\sin \frac{B+C}{2}=\sin A$

$\frac{B+C}{2}=A$

B+C=2A

∴ A+B+C=180°

A+2A=180°

3A=180°

$A=\dfrac{180^{\circ}}{3}=60^{\circ}$


(ix) $(a+b+c)\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)=2 c \cot \frac{C}{2}$
Sol :
$\frac{a+b+c}{2 c}= \frac{\cot \frac{c}{2}}{\tan \frac{A}{2}+\tan \frac{B}{2}}$

R.H.S
$\frac{\cot \frac{c}{2}}{\tan \frac{A}{2}+\tan \frac{B}{2}}=\frac{\frac{s(s-c)}{\Delta}}{\frac{(s-b)(s-c)}{\Delta}+\frac{(s-c)(s-a)}{\Delta}}$

$=\frac{\frac{s(s-c)}{a}}{\frac{(s-b)(s-c)+(s-c)(s-a)}{\Delta}}$

$=\frac{s(s-c)}{(s-c)(s-b+s-a)}$

$=\frac{s}{2 s-b-a}$

$=\frac{a+b+c}{2(a+b+c-b-a)}$

$=\frac{a+b+c}{2 c}$

Question 21

(i) $a \cos \frac{B-C}{2}=(b+c) \sin \frac{A}{2}$
Sol :
$\frac{a}{b+c}=\frac{\sin \cdot \frac{A}{2}}{\cos \frac{b-c}{2}}$

$=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

a=ksinA , b=ksinB  , c=ksinC

L.H.S
$\frac{a}{b+c}=\frac{k \sin A}{k\sin B+k \sin C}$

$=\frac{k \sin A}{k(\sin B+\sin C)}$

$=\frac{\sin [180^{\circ}-(B+C)]}{\sin B+\sin C}$

$=\frac{\sin (B+C)}{\sin B+\sin C}$

$=\frac{2 \sin \frac{B+C}{2}}{2\sin\frac{B+C}{2} \cos \frac{B-C}{2}}$

$=\frac{\cos \left(\frac{180^{\circ}-A}{2}\right)}{\cos \frac{B-C}{2}}$

$=\frac{\cos \left(\frac{180^{\circ}}{2}-\frac{A}{2}\right)}{\cos \frac{B-C}{2}}$

$=\frac{\sin \frac{A}{2}}{\cos \frac{B-C}{2}}$


(ii) $\operatorname{asin}\left(\frac{A}{2}+B\right)=(b+c) \sin \frac{A}{2}$
Sol :
$\frac{a}{b+c}=\frac{\sin A}{\sin \left(\frac{A}{2}+B\right)}$

$\frac{a}{b+c}=\frac{k \sin A}{k\sin B+k \sin C}$

$=\frac{k \sin A}{k(\sin B+\sin C)}$

L.H.S
$\frac{a}{b+c}=\frac{k \sin A}{k \sin B+k \sin C}$

$=\frac{k \sin A}{k(\sin B+\sin C)}$

$=\frac{\sin [180^{\circ}-(B+C)]}{\sin B+\sin C}$

$=\frac{\sin (B+C)}{\sin B+\sin C}$

$=\frac{2 \sin \frac{B+C}{2} \cos \frac{B+C}{2}}{2 \sin \frac{B+C}{2} \cos \frac{B-C}{2}}$

$=\frac{\cos \left(\frac{180^{\circ}-A}{2}\right)}{\cos \left(\frac{B-(180^{\circ}-A-B)}{2}\right)}$

$=\frac{\cos \left(\frac{\sin }{2}-\frac{A}{2}\right)}{\cos \left(\frac{B-180^{\circ} +A+B}{2}\right)}$

$=\frac{\sin \frac{A}{2}}{\cos \left(\frac{-180^{\circ}+A+2 B}{2}\right)}$

$=\frac{\sin \frac{A}{2}}{\cos \left(-\frac{180^{\circ}-A-2 B}{2}\right)}$

$=\frac{\sin \frac{A}{2}}{\cos \left(\frac{180^{\circ}-A-2 B}{2}\right)}$

$=\frac{\sin \frac{A}{2}}{\cos \left( \frac{180}{2}-\frac{A}{2}-\frac{2 B}{2} \right)}$

$=\frac{\sin \frac{A}{2}}{\cos \left[90^{\circ}-\left(\frac{A}{2}+B\right)\right]}$

$=\frac{\sin \frac{A}{2}}{\sin \left(\frac{A}{2}+B\right)}$

Question 22

(i) $a^{2}\left(\cos ^{2} B-\cos ^{2} C\right)+b^{2}\left(\cos ^{2} C-\cos ^{2} A\right)+c^{2}\left(\cos ^{2} A-\cos ^{2} B\right)=0$
Sol :
$=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

a=ksinA , b=ksinB  , c=ksinC

L.H.S
$a^{2}\left(\cos ^{2} B-\cos ^{2} C\right)+b^{2}\left(\cos ^{2} C-\cos ^{2} A\right)+c^{2}\left(\cos ^{2} A-\cos ^{2} B\right)$

$=k^{2} \sin ^{2} A\left(\cos ^{2} B-\cos ^{2} C\right)+k^{2} \sin ^{2} B\left(\cos ^{2}C\left.-\cos ^{2} A\right)+k^{2} \sin ^{2} C\left(\cos ^{2} A-\cos ^{2} B\right)\right.$

$=k^{2}\left(1-\cos ^{2} A\right)\left(\cos ^{2} B-\cos ^{2} C\right)+k^{2}\left(1-\cos ^{2} B\right)\left(\cos ^{2} C-\cos ^{2} A\right)+k^{2}\left(1-\cos ^{2} C\right)\left(\cos ^{2} A-\cos ^{2} B\right)$

$=k^{2}\left[\cos ^{2} B-\cos ^{2} C-\cos ^{2} A \cos ^{2} B+\cos ^{2} A \cos ^{2} C+\cos ^{2} C-\cos ^{2} A\right. -\cos ^{2} B \cos ^{2} C+\cos ^{2} B \cos ^{2} A+\cos ^{2} A-\cos ^{2} B-\cos ^{2} C \cos ^{2} A \left.+\cos ^{2} C \cos ^{2} B \right]$

$=k^{2} \times 0=0$


(ii) $(a+b+c) \sin \frac{A}{2}=2 a \cos \frac{B}{2} \cos \frac{c}{2}$
Sol :
$\frac{a+b+c}{2 a}=\frac{\cos \frac{B}{2} \cos \frac{C}{2}}{\sin \frac{A}{2}}$

R.H.S
$\frac{\cos \frac{B}{2} \cos \frac{C}{2}}{\sin \frac{A}{2}}=\frac{\sqrt{\frac{s(s-b)}{a c}} \times \sqrt{\frac{S(s-c)}{a b}}}{\sqrt{\frac{(s-b)(s-c)}{b c}}}$

$=\frac{s}{a}=\frac{a+b+c}{2 a}$

Question 23

Sol :
A=2B (given)

sinA=sin2B

sinA=2sinBcosB

Let $=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

a=ksinA , b=ksinB  , c=ksinC

$\frac{a}{k}=2 \frac{b}{k} \times \frac{a^{2}+c^{2}-b^{2}}{2 a c}$

$a^{2} c=a^{2} b+b c^{2}-b^{3}$

$0=a^{2} b-a^{2} c-b^{3}+b c^{2}$

$0=a^{2}(b-c)-b\left(b^{2}-c^{2}\right)$

$0=a^{2}(b-c)-b(b-c)(b+c)$

$0=(b-c)\left[a^{2}-b(b+c)\right]$

b-c=0
b=c

$a^{2}-b(b+c)=0$
$a^{2}=b(b+c)$

Question 24

$\frac{\cos A}{a}+\frac{a}{b c}=\frac{\cos B}{b}+\frac{b}{c a}=\frac{\cos C}{c}+\frac{c}{a b}$
Sol :
$\frac{\cos A}{a}+\frac{a}{b c}=\frac{b^{2}+c^{2}-a^{2}}{2 a b c}+\frac{a}{b c}$

$=\frac{b^{2}+c^{2}-a^{2}+2 a^{2}}{2 a b c}$

$=\frac{a^{2}+b^{2}+c^{2}}{2 a b c}$

$\frac{\cos B}{b}+\frac{b}{c a}=\frac{a^{2}+c^{2}-b^{2}}{2 a b c}+\frac{b}{c a}$

$=\frac{a^{2}+c^{2}-b^{2}+2 b^{2}}{2 a b c}$

$=\frac{a^{2}+b^{2}+c^{2}}{2 a b c}$

$\frac{\cos C}{c}+\frac{c}{a b}=\frac{a^{2}+b^{2}-c^{2}}{2 a b c}+\frac{c}{a b}$

$=\frac{a^{2}+b^{2}-c^{2}+2 c^{2}}{2 a b c}$

$=\frac{a^{2}+b^{2}+c^{2}}{2 a b c}$

$\frac{\cos A}{a}+\frac{a}{b c}=\frac{\cos B}{b}+\frac{b}{a c}=\frac{\cos C}{c}+\frac{c}{a b}$

Question 25

(i) $\frac{\cos 2 \mathrm{A}}{a^{2}}-\frac{\cos 2 \mathrm{B}}{b^{2}}=\frac{1}{a^{2}}-\frac{1}{b^{2}}$
Sol :
$=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

a=ksinA , b=ksinB  , c=ksinC

L.H.S
$\frac{\cos 2 A}{a^{2}}-\frac{\cos 2 B}{b^{2}}$

$=\frac{\cos 2 A}{k^{2} \sin ^{2} A}-\frac{\cos 2 B}{k^{2} \sin ^{2} B}$

$=\frac{\left(1-2 \sin ^{2} A\right)}{k^{2} \sin ^{2} A}-\frac{\left(1-2 \sin ^{2} B\right)}{k^{2} \sin^2 B}$

$=\left[\frac{1}{k^{2} \sin ^{2} A}-\frac{2 \sin ^{2} A}{k^{2} \sin ^{2} A}\right)-\left(\frac{1}{k^{2} \sin ^{2} B}-\frac{2 \sin ^{2} B}{k^{2} \sin ^{2} B}\right)$

$=\frac{1}{(k+\sin A)^{2}}-\frac{2}{k^{2}}-\frac{1}{(k \sin B)^{2}}+\frac{2}{k^{2}}$

$=\frac{1}{a^{2}}-\frac{1}{b^{2}}$


(ii) $\left(b^{2}-c^{2}\right) \sin ^{2} A+\left(c^{2}-a^{2}\right) \sin ^{2} B+\left(a^{2}-b^{2}\right) \sin ^{2} C=0$
Sol :
$=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

a=ksinA , b=ksinB  , c=ksinC

L.H.S
$\left(b^{2}-c^{2}\right) \sin ^{2} A+\left(c^{2}-a^{2}\right) \sin ^{2} B+\left(a^{2}-b^{2}\right) \sin ^{2} C$

$\left(k^{2} \sin ^{2} B-k^{2} \sin ^{2} C\right) \sin ^{2} A +( k^{2} \sin ^{2}C-k^2 \sin ^{2} A) \sin ^{2} B+\left(k^{2} \sin ^{2} A-k^{2} \sin ^{2} B\right) \sin ^{2} C$

$\left.=k^{2}\left[\sin ^{2} B \sin ^{2} A-\sin ^{2} C \sin ^{2} A+\sin ^{2} C \sin ^{2} B-\sin ^{2} A \sin ^{2} B\right. +\sin ^{2} A \sin ^{2} C-\sin ^{2} B \sin ^{2} C\right]$

$=k^{2} \times 0=0$


(iii) $\frac{a^{2} \sin (B-C)}{\sin A}+\frac{b^{2} \sin (C-A)}{\sin B}+\frac{c^{2} \sin (A-B)}{\sin C}=0$
Sol :
$=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

a=ksinA , b=ksinB  , c=ksinC

$\frac{a^{2} \sin (B-C)}{\sin A}=\frac{k^{2} \sin ^{2} A \sin (B-C)}{\sin A}$

$=k^{2} \sin [180^{\circ} -(B+C)] \sin (B-C)$

$=k^{2} \sin (B+C) \sin (B-C)$

$\frac{a^{2}-\sin (B-C)}{\sin A}=k^{2}\left[\sin ^{2} B-\sin ^{2} C\right]$..(i)

$b^{2} \frac{\sin (C-A)}{\sin B}=k^{2}\left[\sin ^{2} C-\sin ^{2} A\right]$..(ii)

$\frac{a^{2} \sin (A-C)}{\sin A}+\frac{b^{2} \sin (C-A)}{\sin B}+\frac{c^{2} \sin (A-B)}{\sin C}$

$=k^{2}\left[\sin ^{2} B-\sin ^{2} C+\sin ^{2} C-\sin ^{2} A+\sin^2 A- \sin^2 B\right]$

$=k^{2} \times 0$

Question 26

Sol :
AD is median in ΔABC

BD=CD

In ΔABD
$\sin B=\frac{A D}{B D} \Rightarrow A D=B D \sin B$

In ΔADC
$\frac{C D}{\sin (A-90^{\circ})}=\frac{A D}{\sin C}$

$\frac{B D}{-\sin \left(\sin 90^{\circ}-A\right)}=\frac{B D\sin B}{\sin C}$

$\frac{1}{-\cos A}=\frac{\sin B}{\sin C}$

sinC=-cosAsinB

sinC+cosAsinB=0

sin[180°-(A+B)]+cosAsinB=0

sin(A+B)+cosAsinB=0

sinAcosB+cosAsinB+cosAsinB=0

sinAcosB+2cosAsinB=0

$\frac{\sin A \cos B}{\cos A \cos B}+\frac{2 \cos A \sin B}{\cos A \cos B}=0$

tanA+2tanB=0

Question 27

(i)
Sol :
$\tan \frac{A}{2}, \tan \frac{B}{2}, \tan \frac{C}{2}$.. are in AP

$\tan \frac{B}{2}-\tan \frac{A}{2}=\tan \frac{C}{2}-\tan \frac{B}{2}$

$\frac{\sin \frac{B}{2}}{\cos \frac{B}{2}}-\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}=\frac{\sin \frac{C}{2}}{\cos \frac{C}{2}}-\frac{\sin \frac{B}{2}}{\cos \frac{B}{2}}$

$\frac{\sin \frac{B}{2} \cos \frac{B}{2}-\cos \frac{B}{2} \sin\frac{A}{2}}{\cos \frac{B}{2} \cos \frac{A}{2}}=\frac{\sin \frac{C}{2} \cos \frac{B}{2}-\cos \frac{C}{2} \sin \frac{B}{2}}{\cos \frac{C}{2} \cos \frac{B}{2}}$

$\frac{\sin \left(\frac{B}{2}-\frac{A}{2}\right)}{\cos \frac{C}{2}}=\frac{\sin \left(\frac{C}{2}-\frac{B}{2}\right)}{\cos \frac{C}{2}}$

$\frac{-\sin \left(\frac{A}{2}-\frac{B}{2}\right)}{\cos\left[90^{\circ}-\left(\frac{B}{2}+\frac{C}{2}\right)\right]}=\frac{-\sin \left(\frac{B}{2}-\frac{C}{2}\right)}{\cos \left[90^{\circ}-\left(\frac{A}{2}+\frac{B}{2}\right)\right]}$

$\frac{\sin \left(\frac{A-B}{2}\right)}{\sin \left(\frac{B+C}{2}\right)}=\frac{\sin \left(\frac{B-C}{2}\right)}{\sin \left(\frac{A+B}{2}\right)}$

$2 \sin \left(\frac{A-B}{2}\right) \sin \left(\frac{A+B}{2}\right)=2 \sin \left(\frac{B-C}{2}\right) \sin \left(\frac{B+C}{2}\right)$

cosB-cosA=cosC-cosB

∴ cosA, cosB , cosC are in A.P


(ii)
Sol :
$=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

a=ksinA , b=ksinB  , c=ksinC

∵ $\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}$

sinAsin(B-C)=sinCsin(A-B)

$\sin [180^{\circ}-(B+C)] \sin (B-C)=\sin [180^{\circ}-(A+B)] \sin (A-B)$

sin(B+C)sin(B-C)=sin(A+B)sin(A-B)

$\sin ^{2} B-\sin ^{2} C=\sin ^{2} A-\sin ^{2} B$

$\frac{b^{2}}{k^{2}}-\frac{c^{2}}{k^{2}}=\frac{a^{2}}{k^{2}}-\frac{b^{2}}{k^{2}}$

$\frac{b^{2}-c^{2}}{k^{2}}=\frac{a^{2}-b^{2}}{k^{2}}$

$-\left(c^{2}-b^{2}\right)=-\left(b^{2}-a^{2}\right)$

∴ $a^{2}, b^{2}, c^{2}$ are in A.P


(iii)
Sol :
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=\frac{1}{k}$

ak=sinA , bk=sinB , ck=sinC

∵ sinA , sinB , sinC are in A.P

sinB-sinA=sinC-sinB

bk-ak=ck-bk

k(b-a)=k(c-b)

b+b=a+c

2b=a+c

2b+b=a+b+c

3b=2s

3b+s=2s+s

3b+s=3s

s=3s-3b

s=3(s-b)

$\frac{1}{3}=\frac{s-b}{s}$

$\frac{1}{3}=\tan \frac{A}{2} \tan \frac{C}{2}$

$1=3 \tan \frac{A}{2} \tan \frac{C}{2}$

Question 28

$\cos ^{2} A+\cos ^{2} B+\cos ^{2} C=\cos B \cos C+\cos C \cdot \cos A+\cos A \cos B$
Sol :

$2 \cos ^{2} A+2 \cos ^{2} B-2 \cos ^{2} C=2 \cos B \cos C+2 \cos C \cos A+2 \cos A \cos B$

$\cos^{2} a+\cos^{2} A+\cos^{2} B+\cos^{2} B+\cos^{2} C+\cos^{2} C-2 \cos B \cos C-2\cos C \cos A-2\cos A \cos B=0$

(cos²B+cos²C-2cosBcosC)+(cos²C+cos²A-2cosCcosA)+(cos²A+cos²B-2cosAcosB)=0

(cosB-cosC)²+(cosC-cosA)²+(cosA-cosB)²=0

cosB-cosC=0
cosB=cosC
B=C

cosC-cosA=0
cosC=cosA
C=A

cosA-cosB=0
cosA=cosB
A=B

∴A=B=C

Question 30

Sol :
$\sin C=\frac{\sin A+\sin B}{\cos A+\cos B}$

$\sin [180^{\circ}-(A+B)]=\frac{\sin A+\sin B}{\cos A+\cos B}$

$\sin (A+B)=\frac{2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}}{2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}}$

$2+ \frac{A+B}{2} \cos \frac{A+B}{2=\frac{\sin \frac{A+B}{2}}{\cos \frac{A+B}{2}}$

$\cos ^{2} \frac{A+B}{2}=\frac{1}{2}$

$\cos ^{2} \frac{A+B}{2}=\left(\frac{1}{\sqrt{2}}\right)^{2}$

$\cos ^{2} \frac{A+B}{2}=\cos ^{2} 45^{\circ}$

$\frac{A+B}{2}=45^{\circ}$

A+B=90°


A+B+C=180°
90°+C=180°
C=90°

Question 31

Sol :
$\frac{\cos A}{b}=\frac{\cos B}{a}$

acosA=bcosB

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

a=ksinA, b=ksinB, c=ksinC


ksinAcosA=ksinBcosB

2sinAcosA=2sinBcosB

sin2A=sin2B

sin2A-sin2B=0

$2 \cos \frac{2A+2 B}{2} \sin \frac{2 A-2 B}{2}=0$

cos(A+B)sin(A-B)=0


cos(A+B)=0
cos(A+B)=cos90°
A+B=90°


sin(A-B)=0
sin(A-B)=sin0°
A-B=0
A=B

A+B+C=180°
90°+C=180°
C=90°

Question 34

(i) $b \cos ^{2} \frac{C}{2}+\operatorname{ccos}^{2} \frac{B}{2}=\operatorname{ccos}^{2} \frac{A}{2}+a \cos ^{2} \frac{C}{2}=a \cos ^{2} \frac{B}{2}+b \cos ^{2} \frac{A}{2}=\frac{a+b+c}{2}$
Sol :
$b \cos ^{2} \frac{c}{2}+c \cos ^{2} \frac{B}{2}=b(\sqrt{\frac{s(s-c)}{a b}})^{2}+c(\sqrt{\frac{s(s-b)}{a c}})^{2}$

$=\frac{b s(s-c)}{a b}+\frac{c s(s-b)}{ac}$

$=\frac{s(s-c)+s(s-b)}{a}$

$=\frac{a[s-c+s-b]}{a}$

$=\frac{s[2s-c-b]}{a}$

$=\frac{5[a+b+c-c-b]}{a}$

$=\frac{5 \times a}{a}=s=\frac{a+b+c}{2}$

$c \cos ^{2} \frac{A}{2}+a \cos ^{2} \frac{C}{2}=C\left.\left(\sqrt{\left.\frac{s(s-a)}{b c}\right.}\right)^{2}+a\left(\sqrt{\frac{s(s-c)}{a b}}\right)^{2}\right.$

$=\frac{c \times s(s-a)}{bc}+a\times \frac{s(s-c)}{a b}$

$=\frac{s(s-a)+s(s-c)}{b}$

$=\frac{s[s-a+s-c]}{b}$

$=\frac{s[2s-a-c]}{b}$

$=\frac{s(a+b+c-a-c)}{b}$

$=\frac{s \times b}{b}=s$

$=\frac{a+b+c}{2}$


$a cos^{2} \frac{B}{2}+b \cos ^{2} \frac{A}{2}=a\left(\sqrt{\frac{s(s-b)}{a c}}\right)^{2}+b\left(\sqrt{\frac{s(s-a)}{b c}}\right)^{2}$

$=a\frac{s(s-b)}{a c}+b \times \frac{s(s-a)}{b c}$

$=\frac{s(s-b)+s(s-a)}{c}$

$=\frac{s[s-b+s-a]}{c}$

$=\frac{s[2s-b-a]}{c}$

$=\frac{s(a+b+c-b-a]}{c}$

$=\frac{s \times c}{c}=s=\frac{a+b+c}{2}$


(ii) $\frac{b-c}{a} \cos ^{2} \frac{A}{2}+\frac{c-a}{b} \cos ^{2} \frac{B}{2}+\frac{a-b}{c} \cos ^{2} \frac{C}{2}=0$
Sol :
L.H.S
$\frac{b-c}{a} \cos ^{2} \frac{A}{2}+\frac{c-a}{b} \cos ^{2} \frac{B}{2}+\frac{a-b}{c} \cos ^{2} \frac{C}{2}$

$=\frac{b-c}{a} \times\left( \frac{s(s-a)}{b c}\right)^{2}+\frac{c-a}{b} \times\left(\sqrt{\frac{s(s-b)}{a c}}\right)^{2}+\frac{a-b}{c} \times\left(\sqrt{\frac{s(s-c)}{a b}}\right)^{2}$

$=\frac{(b-c) \cdot s(s-a)}{a b c}+\frac{(c-a) \cdot s(s-b)}{a b c}+\frac{(a-b) \cdot s(s-c)}{a b c}$

$=\frac{s}{a b c}[(b-c)(s-a)+(c-a)(s-b)+(a-b)(s-c)]$

$=\frac{S}{a b c}[5 b-a b-s c+c a+s c-b c-5 a+a b+s a-c q$

$=\frac{s}{a b c} \times 0=0$


(iii) $(b-c) \cot \frac{A}{2}+(c-a) \cot \frac{B}{2}+(a-b) \cot \frac{C}{2}=0$
Sol :
L.H.S
$(b-c) \cot \frac{A}{2}+(c-a) \cot \frac{B}{2}+(a-b) \cot \frac{C}{2}$

$=(b-c) \cdot \frac{s(s-a)}{\Delta}+(c-a) \frac{s(s-b)}{\Delta}+(a-b) \cdot \frac{s(s-c)}{\Delta}$

$=\frac{5}{\Delta}[(b-c)(s-a)+(c-a)(s-b)+(a-b)(s-c)]$

$=\frac{S}{\Delta} \times 0=0$

Question 35

Sol :
$a\cos ^{2} \frac{C}{2}+\operatorname{ccos}^{2} \frac{A}{2}=\frac{3 b}{2}$

$a\left(\sqrt{\frac{5(s-2)}{a b}}\right)^{2}+c\left(\sqrt{\frac{s(s-a)}{b c}}\right)^{2}=\frac{3 b}{3}$

$a \times \frac{s(s-c)}{a b}+c \times \frac{s(s-a)}{b c}=\frac{3 b}{2}$

$\frac{s(s-c)+s(s-a)}{b}=\frac{3 b}{2}$

$\frac{s[s-c+s-a]}{b}=\frac{3 b}{2}$

$\frac{s[2s-c-a]}{b}=\frac{3 b}{2}$

$\frac{s[a+b+c-c-a]}{b}=\frac{3 b}{2}$

$\frac{s \times b}{b}=\frac{3b}{2}$

2s=3b

a+b+c=3b

a+c=2b

or c-b=b-a

∴ a ,b,c are in A.P

Question 36

(i) $(a+b+c)\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)=2 \cot \frac{C}{2}$
Sol :
$\frac{a+b+c}{2}=\frac{\cot \frac{c}{2}}{\tan \frac{A}{2}+tan\frac{B}{2}}$

R.H.S
$\frac{\cot \frac{C}{2}}{\tan \frac{A}{2}+\tan \frac{B}{2}}=\frac{\frac{s(s-c)}{\Delta}}{\frac{(s-b)(s-c)}{\Delta}+\frac{(s-c)(s-a)}{\Delta}}$

$=\frac{\frac{s(s-c)}{\Delta}}{\frac{(s-b)(s-c)+(s-c)(s-q)}{\Delta}}$

$=\frac{s(s-c)}{(s-c)[s b+s-a]}$

$=\frac{s}{[2s-b-a]}$

$=\frac{5}{(a+b+c-b-a)}$

$=\frac{a+b+c}{2}$


(ii) $1-\tan \frac{\mathrm{A}}{2} \tan \frac{\mathrm{B}}{2}=\frac{2 c}{a+b+c}$
Sol :
$1-\tan \frac{A}{2} \tan \frac{B}{2}$

$=1-\frac{(s-b)(s-c)}{s} \times \frac{(s-c)(s-a)}{\Delta}$

$=\frac{\Delta^{2}-(s-a)(s-b)(s-c)^{2}}{\Delta^{2}}$

[∵$\delta=\sqrt{s(s-a)(s-b)(s-c)}$]

$=\dfrac{s(s-a)(s-b)(s-c)-\frac{s(s-a)(s-b)(s-c)^{2}}{2}}{s(s-a)(s-b)(s-c)}$

$=\dfrac{s(s-a)(s-b)(s-c)\left[1-\frac{s(s-c)}{s}\right]}{s(s-a)(s-b)(s-c)}$

$=1-\frac{s}{s}+\frac{c}{s}$

$=1-1+\dfrac{C}{\frac{a+b+c}{2}}$

$=\dfrac{2 C}{a+b+c}$


(iii) $1-\tan \frac{\mathrm{A}}{2} \tan \frac{\mathrm{B}}{2}=\frac{2 c}{a+b+c}$
Sol :
L.H.S
$1-\tan \frac{A}{2} \tan \frac{B}{2}$

[∵ $\tan \frac{A}{2}\tan\frac{B}{2}=\frac{s-c}{s}$]

$=1-\frac{s-c}{s}$

$=1-\frac{s}{s}+\frac{c}{s}$

$=1+1+\frac{c}{\frac{a+b+c}{2}}=\frac{2 c}{a+b+c}$

Question 37

(i) $(b+c-a) \tan \frac{A}{2}=(c+a-b) \tan \frac{B}{2}=(a+b-c) \tan \frac{C}{2}$
Sol :
$(b+c-a) \tan \frac{A}{2}=(a+b+c-2 a) \frac{(s-b)(s-c)}{\Delta}$

$=\frac{(2s-2 a)(s-b)(s-c)}{\Delta}$

$=\frac{2(s-a)(s-b)(s-c)}{\Delta}$

$(c+a-b) \tan \frac{B}{2}=(a+b+c-2 b)\frac{(s-c)(s-a)}{\Delta}$

$=\dfrac{(2s-2 b)(s-c)(s-a)}{\Delta}$

$=\frac{2(s-b)(s-c)(s-a)}{\Delta}$

$=2\frac{(s-a)(s-b)(s-c)}{2}$


$(a+b-c) \tan \frac{c}{2}=(a+b+c-2 c) \frac{(s-a)(s-b)}{\Delta}$

$=\frac{(2s-2 c)(s-a)(s-b)}{\Delta}$

$=\frac{2(s-c)(s-a)(s-b)}{\Delta}$


(ii) $\frac{\cos ^{2} \frac{A}{2}}{a}+\frac{\cos ^{2} \frac{C}{2}}{b}+\frac{\cos ^{2} \frac{C}{2}}{c}=\frac{s^{2}}{a b c}$
Sol :
L.H.S
$\frac{\cos ^{2} \frac{A}{2}}{a}+\frac{\cos ^{2} \frac{B}{2}}{b}+\frac{\cos ^{2} \frac{C}{2}}{c}$

$=\dfrac{\left(\sqrt{\left.\frac{s(s-a)}{b c}\right.}\right)^2}{a}+\dfrac{\left(\sqrt{\frac{s(s-b)}{a c}}\right)^2}{b}+\dfrac{\left(\sqrt{\frac{s(s-c)}{a b}}\right)^{2}}{c}$

$=\frac{s(s-a)}{a b c}+\frac{s(s-b)}{a b c}+\frac{s(s-c)}{a b c}$

$=\frac{s(s-a)+s(s-b)+s(s-c)}{a b c}$

$=\frac{s[s-a+s-b+s-c]}{a b c}$

$=\frac{s[3 s-(a+b+c)]}{a b c}$

$=\frac{s[3 s-2 s]}{a b c}$

$=\frac{s \times s}{a b c}$

$=\frac{s^{2}}{a b c}$

Question 38

(i) $\left(\cot \frac{A}{2}+\cot \frac{B}{2}\right)\left(a \sin ^{2} \frac{B}{2}+b \sin ^{2} \frac{A}{2}\right)=\operatorname{ccot} \frac{C}{2}$
Sol :
L.H.S
$\left(\cot\frac{A}{2}+\cot\frac{B}{2}\right)\left(a\sin^2 \frac{B}{2}+b \sin ^{2} \frac{A}{2}\right)$

$=\left[\frac{s(s-a)}{\Delta}+\frac{s(s-b]}{\Delta}\right]\left[a\left(\sqrt{\frac{(s-c)(s-a)}{c a}}\right)^{2}+b\left(\sqrt{\frac{(s-b)(s-c)}{b c}}\right)^{2}\right]$

$=\frac{s}{\Delta}[s-a+s-b)\left[\begin{array}{cccc}a \frac{(s-c)(s-a)}{c a} +b\left(\frac{s-b)(s-c)}{b c}\right.\end{array}\right]$

$=\frac{s}{\Delta}(2s-a-b) \frac{(s-c)}{c}[s-a+s-b]$

$=\frac{S}{\Delta}\left(a+b+c-a-b\right) \frac{(s-c)}{c}(a+b+c-a-b)$

$=\frac{s}{\Delta} \times c \times \frac{(s-c)}{c} \times c$

$=c \cdot \frac{s(s-c)}{\Delta}=c \cdot \cot \frac{c}{2}$

(ii) $\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}=\sqrt{\left(1-\frac{a}{s}\right)\left(1-\frac{b}{s}\right)\left(1-\frac{c}{s}\right)}$
Sol :
L.H.S
$\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}$

$=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \times \sqrt{\frac{(s-c)(s-a)}{s(s-b)}} \times \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$

$=\sqrt{\left(\frac{s}{s}-\frac{c}{s}\right)} \times \sqrt{\left(\frac{s-a}{s}-\frac{a}{s}\right)} \times \sqrt{\left(\frac{s}{s}-\frac{b}{s}\right)}$

$=\sqrt{\left(1-\frac{a}{s}\right)\left(1-\frac{b}{s}\right)\left(1-\frac{c}{s}\right)}$

Question 40

Sol :
[]
𝛼-β+𝛼+𝛼+β=3×side

3𝛼=3×side

𝛼=side


[]
$\sqrt{s(s-a)(s-b)(s-c)}=\frac{3}{s} \times \frac{\sqrt{3}}{4}(side)^2$

$\sqrt{\frac{3 \alpha}{2}\left[\frac{3 \alpha}{2}-(\alpha-\beta)\right]\left[\frac{3 \alpha}{2}-\alpha\right]\left[\frac{3 \alpha}{2}-(\alpha+\beta)\right.}=\frac{3}{5} \times \frac{\sqrt{3}}{4} \alpha^{2}$

$\sqrt{\frac{3 \alpha}{2} \times \left(\frac{\alpha}{2}+\beta\right) \cdot \frac{\alpha}{2} \times\left(\frac{\alpha}{2}-\beta\right)}=\frac{3}{5} \times \frac{\sqrt{3}}{4} \alpha^{2}$

$\sqrt{\frac{3 \alpha}{2} \times\left(\frac{\alpha+2 \beta}{2}\right) \times \frac{\alpha}{2} \times\left(\frac{\alpha-2 \beta}{2}\right)}=\frac{3}{5} \times \frac{\sqrt{3}}{4} \alpha^{2}$

$\frac{\alpha}{2} \times \frac{1}{2} \sqrt{3(\alpha+2 \beta)(\alpha-2 \beta)}=\frac{3}{5} \times \frac{\sqrt{3}}{4} \alpha^{2}$

$\frac{\alpha}{4} \sqrt{3} \times \sqrt{(\alpha+2 \beta)(\alpha-2 \beta)}=\frac{3}{5} \times \frac{\sqrt{3}}{4} \alpha ^2$

$\sqrt{\alpha^{2}-(2 \beta)^{2}}=\frac{3}{5} \alpha$

[]

$(\sqrt{\alpha^{2}-4 \beta^{2}})^{2}=\left(\frac{3}{5} \alpha\right)^{2}$

$\alpha^{2}-4 \beta^{2}=\frac{9 \alpha^{2}}{25}$

$25 \alpha^{2}-100 \beta^{2}=9 \alpha^{2}$

$16 \alpha^{2}=100 \beta^{2}$

$\frac{\alpha^{2}}{\beta^{2}}=\frac{100}{16}$

$\frac{\alpha}{\beta}=\sqrt{\frac{25}{4}}=\frac{5}{2}$

let 𝛼=5k ,β=2k


∴

First side=𝛼-β=5k-2k=3k

second side=𝛼=5k

third side=𝛼+β=5k+2k=7k

∴
ratios of sides 3:5:7

Question 41

Sol :
a=3m ,b=4m , c=5m , d=6m

A+C=120°

$s=\frac{a+b+c+d}{2}=\frac{3+4+5+6}{2}$

$=\dfrac{18}{2}=9cm$


$=\sqrt{(s-a)(s-b)(s-c)(s-d)-a b c d \cos ^{2}\left(\frac{A+C}{2}\right)}$

$=\sqrt{(9-3)(9-4)(9-5)(9-6)-3 \times 4 \times 5 \times 6 \cos ^{2} \frac{120^{\circ}}{2}}$

$=\sqrt{6 \times 5 \times 4 \times 3-3 \times 4 \times 5 \times 6 \times\left(\frac{1}{2}\right)^{2}}$

$=\sqrt{3 \times 4 \times 5 \times 6\left(1-\frac{1}{4}\right)}$

$=\sqrt{3 \times 2 \times 2 \times 5 \times 3 \times 2 \times \frac{3}{4}}$

$=3 \sqrt{2 \times 5 \times 3}$

$=3 \sqrt{30} \mathrm{cm}^{2}$

Question 42

$r_{1}+r_{2}+r_{3}-r=4 R$
Sol :
L.H.S
$r_{1}+r_{2}+r_{3}-r$

$=\frac{\Delta}{s-a}+\frac{\Delta}{s-b}+\frac{\Delta}{s-c}-\frac{\Delta}{s}$

$\left.=\frac{\Delta[s(s-b)(s-c)+s(s-a)(s-c)+s(s-a)(s-b)-(s-a)(s-b)(s-c)}{s(s-a)(s-b)(s-c)}\right]$

[∵ $\Delta=\sqrt{s(s-a)(s-b)(s -c)}$ ]

$=\frac{\Delta}{\Delta^{2}}\left[\begin{array}{c}s\left(s^{2}+s c-s b+b c\right)+s\left(s^{2}-s c-s a+a c\right)+s\left(s^{2}-s b-s a+a b\right) -(s-a)\left(s^{2}-s c-s b+b c\right)\end{array}\right]$

$\begin{aligned} {=\frac{1}{\Delta}\left[s^{3}-s^{2} c-s^{2} b+s b c+s^{3}-s^{2} c-s^{2} a+s a c+s^{3}-s^{2} b\right.}{-s^{2} a+s a b-s^{3}+s^{2} c+s^{2} b-s b c+s^{2} a-s a c}-s a b+a b c] \end{aligned}$

$=\frac{1}{\Delta}\left[2 s^{3}-s^{2} c-s^{2} b-s^{2} a+a b c\right]$

$=\frac{1}{\Delta}\left[2s^{3}-s^{2}(c+b+a)+a b c\right]$

$=\frac{1}{\Delta}\left[2 s^{3}-s^{2} \cdot 2 s+a b c\right]$

$=\frac{1}{\Delta}\left[2 s^{3}-2 s^{3}+a b c\right]$

$=\frac{a b c}{\Delta}=4R$