KC Sinha Mathematics Solution Class 11 Chapter 6 त्रिकोणमितीय फलन (Trigonometric function) Exercise 6.1

Exercise 6.1

Question 1

साबित करे (prove that)
(i) $\sin 15^{\circ}=\frac{\sqrt{3}-1}{2 \sqrt{2}}$
Sol :
L.H.S
sin15°=sin(45°-30°)

[∵sin(A-B)=sinAcosB-cosAsinB]

$=\sin 45^{\circ} \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ}$

$=\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2}$

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

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


(ii)$\cos 75^{\circ}=\frac{\sqrt{3}-1}{2 \sqrt{2}}$
Sol :
L.H.S
$\cos 75^{\circ}=\cos (45^{\circ}+30^{\circ})$


(iii)$\tan 75^{\circ}=2+\sqrt{3}$
Sol :
L.H.S
$\tan 75^{\circ}=\tan (45^{\circ}+30^{\circ})$

$\tan (A+B)=\frac{\tan A+tanB}{1-\tan A.tanB}$

$=\frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 40^{\circ} \tan 30^{\circ}}$

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

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

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

$=\frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

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

$=\frac{4+2 \sqrt{3}}{2}$

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

$\tan 75^{\circ}=2+\sqrt{3}$


(iv)$\tan 15^{\circ}=2-\sqrt{3}$
Sol :

(v)$\sin 75^{\circ}=\frac{\sqrt{3}+1}{2 \sqrt{2}}$

Question 2

निम्नलिखित के मान निकाले
Find the value of the following :
(i) $\tan \left(\frac{11 \pi}{12}\right)$
Sol :
$\tan \left(\frac{11 \pi}{12}\right)=\tan \left(\pi-\frac{\pi}{12}\right)$

$=-\tan \frac{\pi}{12}$

$=-\tan \left(\frac{\pi}{4}-\frac{\pi}{6}\right)$
[$\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \cdot \tan B}$]

$=-\frac{\tan \frac{\pi}{4}-\tan \frac{\pi}{6}}{1+\tan \frac{\pi}{4} \times \tan \frac{\pi}{6}}$

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

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

$=-\frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}$

$=-\left(\frac{3-\sqrt{3}-\sqrt{3}+1}{(\sqrt{3})^{2}-1^{2}}\right]$

$=-\left(\frac{3-\sqrt{3}-\sqrt{3}+1}{(\sqrt{3})^{2}-1^{2}}\right]$

$=-\left[\frac{4-2 \sqrt{3}}{2}\right]$

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

$=-2+\sqrt{3}=\sqrt{3}-2$


(ii) $\tan \left\{(-1)^{n} \frac{\pi}{4}\right\}$ where n is integer
Sol :
[tan(-θ)=tanθ

$\tan \left(\frac{\pi}{4}\right)=1$

$\tan \left(-\frac{\pi}{4}\right)=-1$]

$\tan \left\{(-1)^{n} \frac{\pi}{4}\right\}$

$=(-1)^{n}\cdot 1$


(iii) $\tan \frac{13 \pi}{12}$
Sol :
$=\tan \left(\pi+\frac{\pi}{12}\right)$

$={\tan }\frac{\pi}{12}$

$=\tan \left(\frac{\pi}{4}-\frac{\pi}{6}\right)$

Question 3

साबित करे(prove that)
(i) sin(n+1)xsin(n+2)x+cos(n+1)xcos(n+2)x=cosx
Sol :
(i)
L.H.S
$\cos (n+1) x \cos (n+2) x+\sin (n+1) x \sin (n+2) x$

$\cos (A-B)=\cos A \cos B+\sin A \sin B$

$=\cos [(n+1) x-(n+2) x]$

$=\cos [n x+x-nx-2 x]$

=cos(-x)

=cosx


(iii) $\cos \left(\frac{\pi}{4}-x\right)\cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right)\sin \left(\sin \frac{\pi}{4}-y\right)=sin(x+y)$
Sol :
$\cos \left(\frac{\pi}{4}-x\right) \cdot \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)$

$\cos (A+B)=cosA \cos B-\sin A \sin B$

$=\cos \left(\frac{\pi}{4}-x+\frac{\pi}{4}-y\right)$

$=\cos \left(\frac{\pi}{2}-x-y\right)$

$=\cos \left[\frac{\pi}{2}-(x+y)\right]$

=sin(x+y)


(iv) tan3x-tan2x-tanx=tan3x.tan2x.tanx
Sol :
3x=2x+x

दोनो तरफ tan लेने पर
(multiplying both sides by tan)

tan 3x=tan (2x+x)

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

$\tan 3 x=\frac{\tan 2 x+\tan x}{1-\tan 2 x \tan x}$

tan3x-tan3x.tan2x.tanx=tan2x+tanx

tan3x-tan2x-tanx=tan3x.tan2x.tanx


(v) $\frac{\tan (\theta+\phi)+\tan (\theta-\phi)}{1-\tan (\theta+\phi) \tan (\theta-\phi)}=\tan 2 \theta$
Sol :
L.H.S
$\frac{\tan (\theta+\phi)+\tan (\theta-\phi)}{1-\tan (\theta+\phi) \tan (\theta-\phi)}$

$=\tan (\theta+\phi+\theta-\phi)$

=tan2θ

Question 4

साबित करे
(Prove that)
(i) $\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x$
Sol :
L.H.S
$\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)$

$=\left[\cos \frac{3 \pi}{4} \cos x-\sin \frac{3 \pi}{4} \sin x\right]-\left[\cos \frac{3 \pi}{4} \cos x+\sin \frac{3 \pi}{4} \sin x\right]$

$=\cos \frac{3 \pi}{4} \cos x-\sin \frac{3 \pi}{4} \sin x-\cos \frac{3 \pi}{4} \cos x-\sin \frac{3 \pi}{4} \sin x$

$=-2 \sin \frac{3 \pi}{4} \sin x$

$=-2 \sin \left(\pi-\frac{\pi}{4}\right) \sin x$

$=-2\left(\sin \frac{\pi}{4}\right) \sin x$

$=-2 \times{\frac{1}{\sqrt{2}}} \sin x$

$=-\sqrt{2} \sin x$


(iii) $\sin ^{2} 6 x-\sin ^{2} 4 x=\sin 2 x \sin 10 x$
Sol :
L.H.S

$\sin ^{2} 6 x-\sin ^{2} 4 x$

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

$=\sin (6 x+4 x) \cdot \sin (6 x-4 x)$

$= \sin 10 x \cdot \sin 2 x$

Question 5

साबित करे
Prove that:
$\cos 9^{\circ}+\sin 9^{\circ}=\sqrt{2} \sin 54^{\circ}$
Sol :
$\cos 9^{\circ}+\sin 9^{\circ}$

$=\sqrt{2} \times \frac{1}{\sqrt{2}}(\cos 9+\sin 9)$

$=\sqrt{2}\left(\frac{1}{\sqrt{2}} \cos 9^{\circ}+\frac{1}{\sqrt{2}} \sin 9^{\circ}\right)$

$=\sqrt{2}(\sin 45^{\circ} \cos 9^{\circ}+\cos 45^{\circ} \sin 9^{\circ}]$

$=\sqrt{2} \sin (45^{\circ}+9^{\circ})$

$=\sqrt{2} \sin 54^{\circ}$

Question 6

साबित करे
Prove that:
$\frac{\cos 20^{\circ}-\sin 20^{\circ}}{\cos 20^{\circ}+\sin 20^{\circ}}=\tan 25^{\circ}$
Sol :
[$\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$]

L.H.S
$\frac{\cos 20^{\circ}-\sin 20^{\circ}}{\cos 20^{\circ}+\sin 20^{\circ}}$

Dividing both sides by $\cos 20^{\circ}$

$=\frac{\frac{\cos 20^{\circ}}{\sin 20^{\circ}}-\frac{\sin 20^{\circ}}{\cos 20^{\circ}}}{\frac{ \cos 20^{\circ}}{\cos 20^{\circ}}+\frac{\operatorname{sin} 20^{\circ}}{\cos 20^{\circ}}}$

$=\frac{1-\tan 20^{\circ}}{1+\tan 20^{\circ}}$

$=\frac{\tan 45^{\circ}-\tan 20^{\circ}}{1+\tan 45^{\circ} \cdot \tan 20^{\circ}}$

$=\tan (45^{\circ}-20^{\circ})=\tan 25^{\circ}$

Question 7

साबित करे
Prove that :
$\frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4}-x\right)}=\left(\frac{1+\tan x}{1-\tan x}\right)^{2}$
Sol :
L.H.S
$\frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4}-x\right)}$

=$\frac{\frac{\tan \frac{\pi}{4}+\tan x}{1-\tan \frac{\pi}{4} \cdot \tan n}}{\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4}\cdot \tan x} }$

$=\frac{\frac{1+\tan x}{1-\tan x}}{\frac{1-\tan x}{1+\tan x}}$

$=\frac{(1+\tan x)^{2}}{(1-\tan x)^{2}}$

$=\left(\frac{1+\tan x}{1-\tan x}\right)^{2}$

Question 8

साबित करे
Prove that:
(i) $\frac{1}{\tan 3 A-\tan A}-\frac{1}{\cot 3 A-\cot A}=\cot 2 A$
Sol :
L.H.S
$\frac{1}{\tan 3 A-\tan A}-\frac{1}{\cot 3 A-\cot A}$

$=\frac{1}{\frac{\sin 3 A}{\cos 3 A}-\frac{\sin A}{\cos A}}-\frac{1}{\frac{\cos 3 A}{\sin 3 A}-\frac{\cos A}{\sin A}}$

$=\frac{1}{\frac{\sin 3 A \cos A-\cos 3 A \sin A}{\cos 3 A \cos A}}-\frac{1}{\frac{\cos 3 A \sin A-\sin 3 A \cos A}{\sin 3 A \sin A}}$

$=\frac{\cos 3 A \cos A}{\sin (3 A-A)}-\frac{\sin 3A \sin A}{\sin (A-3 A)}$

$=\frac{\cos 3 A \cos A}{\sin 2 A}-\frac{\sin 3 A \sin A}{\sin (-2 A)}$

$=\frac{\cos 3 A \cos A}{\sin 2 A}+\frac{\sin 3 A \sin A}{\sin 2 A}$

$=\frac{\cos 3 A \cos A+\sin 3 A \sin A}{\sin 2 A}$

$=\frac{\cos (3 A-A)}{\sin 2 A}$

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


(iii) $\frac{\sin 3 \alpha}{\sin \alpha}+\frac{\cos 3 \alpha}{\cos \alpha}=4 \cos 2 \alpha$
Sol :
L.H.S
$\frac{\sin 3 \alpha}{\sin \alpha}+\frac{\cos 3 \alpha}{\cos \alpha}$

$=\frac{\sin 3 \alpha \cos \alpha+\cos 3 \alpha \sin \alpha}{\sin \alpha \cos \alpha}$

$=\frac{\sin (3 \alpha+\alpha)}{\sin \alpha \cos \alpha}$

$=\frac{\sin 4 \alpha}{\sin \alpha \cos \alpha}$
[sin2A=2sinAcosA]

$=\frac{2 \sin 2(2 \alpha)}{2 \sin \alpha \cos \alpha}$

$=\frac{2 \times 2 \sin 2 \alpha \cos 2 \alpha}{\sin 2 \alpha}$\

$=4 \cos 2 \alpha$

Question 9

साबित करे
Prove that :
$\frac{\tan \left(\frac{\pi}{4}+\mathrm{A}\right)-\tan \left(\frac{\pi}{4}-\mathrm{A}\right)}{\tan \left(\frac{\pi}{4}+\mathrm{A}\right)+\tan \left(\frac{\pi}{4}-\mathrm{A}\right)}=\sin 2 \mathrm{A}$
Sol :
L.H.S
$\frac{\tan \left(\frac{\pi}{4}+\mathrm{A}\right)-\tan \left(\frac{\pi}{4}-\mathrm{A}\right)}{\tan \left(\frac{\pi}{4}+\mathrm{A}\right)+\tan \left(\frac{\pi}{4}-\mathrm{A}\right)}$

$=\frac{\frac{\operatorname{sim}\left(\frac{\pi}{4}+A\right)}{\cos \left(\frac{\pi}{4}+A\right)}-\frac{\sin \left(\frac{\pi}{4}-A\right)}{\cos \left(\frac{\pi}{4}-A\right)}}{\frac{\sin \left(\frac{\pi}{4}+A\right)}{\cos \left(\frac{\pi}{4}+A\right)}+\frac{\sin \left(\frac{\pi}{4}-A\right)}{\cos \left(\frac{\pi}{4}-A\right)}}$

$=\dfrac{\frac{\sin \left(\frac{\pi}{4}+A\right) \cos \left(\frac{\pi}{4}-A\right)-\cos \left(\frac{\pi}{4}+A\right) \cdot \sin \left(\frac{\pi}{4}-A\right)}{\cos \left(\frac{\pi}{4}+A\right) \cos \left(\frac{\pi}{4}-A\right)}}{\frac{\sin \left(\frac{\pi}{4}+A\right) \cos \left(\frac{\pi}{4}-A\right)+\cos \left(\frac{\pi}{4}+A\right) \sin \left(\frac{\pi}{4}-A\right)}{\cos \left(\frac{\pi}{4}+A\right) \cos \left(\frac{\pi}{4}-A\right)}}$

$=\frac{\sin \left[\left(\frac{\pi}{4}+A\right)-\left(\frac{\pi}{4}-A\right)\right]}{\sin \left[\left(\frac{\pi}{4}+A\right)+\left(\frac{\pi}{4}-A\right)\right]}$

$=\frac{\sin \left(\frac{\pi}{4}+A-\frac{\pi}{4}+A\right)}{\sin \left(\frac{\pi}{4}+A+\frac{\pi}{4}-A\right)}$

$=\frac{\sin 2 A}{\sin \frac{\pi}{2}}=\sin 2 A$

Question 10

साबित करे
Prove that:
$\tan 40^{\circ}+2 \tan 10^{\circ}=\tan 50^{\circ}$
Sol :
50°=40°+10°

tan50°=tan(40°+10°)

$\tan 50^{\circ}=\frac{\tan 40^{\circ}+\tan 10^{\circ}}{1-\tan 40^{\circ} \tan 10^{\circ}}$

tan50°-tan50°tan40°tan10°=tan40°+tan10°

tan50°-cot(90°-50°)tan40°tan10°=tan40°+tan10°

tan50°-cot40°tan40°tan10°=tan40°+tan10°

tan50°-tan10°=tan40°+tan10°

tan50°=tan40°+2tan10°

Question 11

साबित करे
Prove that:
$\tan (\alpha+\beta)\tan (\alpha-\beta)=\frac{\sin ^{2} \alpha-\sin ^{2} \beta}{\cos ^{2} \alpha-\sin ^{2} \beta}$
Sol :
[sin(A+B)sin(A-B)$=\sin ^{2} A-\sin ^{2} B$

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

R.H.S
$\frac{\sin ^{2} \alpha-\sin ^{2} \beta}{\cos ^{2} \alpha-\sin ^{2} \beta}$

$=\frac{\sin (\alpha+\beta) \sin (\alpha-\beta)}{\operatorname{cos}(\alpha+\beta) \cos \beta(\alpha-\beta)}$

$=\tan (\alpha+\beta) \tan (\alpha-\beta)$

Question 12

साबित करे
Prove that :
$\tan ^{2} \alpha-\tan ^{2} \beta=\frac{\sin (\alpha+\beta) \sin (\alpha-\beta)}{\cos ^{2} \alpha \cos ^{2} \beta}$
Sol :
[sin(A+B)sin(A-B)$=\sin ^{2} A-\sin ^{2} B$]

R.H.S
$=\frac{\sin (\alpha+\beta) \sin (\alpha-\beta)}{\cos ^{2} \alpha \cos ^{2} \beta}$

$=\frac{\cos ^{2} \beta-\cos ^{2} \alpha}{\cos ^{2} \alpha \cos ^{2} \beta}$

$=\frac{\cos ^{2} \beta }{\cos ^{2} \alpha \cos ^{2} \beta}-\frac{\cos ^{2} \alpha}{\cos ^{2} \alpha \cos ^{2} \beta}$

$=\sec ^{2} \alpha-\sec ^{2} \beta$

$=\left(1+\tan ^{2} \alpha\right)-\left(1+\tan ^{2} \beta\right)$

$=1+\tan ^{2} \alpha-1-\tan ^{2} \beta$

$=\tan ^{2} \alpha-\tan ^{2} \beta$

Question 13

साबित करे
Prove that:
(i) $\tan \{(2 n+1) \pi+\theta\}+\tan \{(2 n+1) \pi-\theta\}=0$
Sol :
L.H.S
$\tan \{(2 n+1) \pi+\theta\}+\tan \{(2 n+1) \pi-\theta\}$

$=\tan \theta+(-\tan \theta)$

=0

(ii) $\tan \left(\frac{\pi}{4}+\theta\right) \tan \left(\frac{3 \pi}{4}+\theta\right)+1=0$
Sol :
L.H.S
$\tan \left(\frac{\pi}{4}+\theta\right) \tan \left(\frac{3 \pi}{4}+\theta\right)+1$

$=\tan \left(\frac{\pi}{4}+\theta\right) \tan \left[\pi-\left(\frac{\pi}{4}-\theta\right)\right]+1$

$=\tan \left(\frac{\pi}{4}+\theta\right)\left[-\tan \left(\frac{\pi}{4}-\theta\right)\right]+1$

$=-\tan \left(\frac{\pi}{4}+\theta\right) \tan \left(\frac{\pi}{4}-\theta\right)+1$

$=-\left(\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4} \cdot \tan \theta} \times \frac{\tan \frac{\pi}{4}-\tan \theta}{1+\tan \frac{\pi}{2} \cdot \tan \theta}\right)+1$

$=\left[-\frac{1+\tan \theta}{1-\tan \theta} \times \frac{1-\tan \theta}{1+\tan \theta}\right]+1$

=-1+1

=0

Question 14

यदि(If) $\tan \alpha=\frac{m}{m+1}, \tan \beta=\frac{1}{2 m+1}$साबित करे (prove that)
$\alpha+\beta=\frac{\pi}{4}$
Sol :
$\tan (\alpha+B)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan B}$

$\tan (\alpha \rightarrow B)=\frac{\frac{m}{m+1}+\frac{1}{2 m+1}}{1-\frac{m}{m+1} \times \frac{1}{2 m+1}}$

$\tan (\alpha+\beta)=\frac{\frac{m(2 m+1)+m+1}{(m+1)(2 m+1)}}{\frac{(m+1)(2 m+1)-m}{(m+1)(2 m+1)}}$

$\tan (\alpha+\beta)=\frac{2 m^{2}+2 m+1}{2 m^{2}+2 m+1}$

$\tan (\alpha+\beta)=\tan \frac{\pi}{4}$

$\alpha+\beta=\frac{\pi}{4}$

Question 15

यदि(If) A+B=45° , साबित करे कि
Show that
(cotA-1)(cotB-1)=2
Sol :
A+B=45°
Taking cot both sides

cot(A+B)=cot45°

$\frac{\cot A \cot B -1}{\cot A+\cot B}=1$

cotAcotB-1=cotA+cotB

cotAcotB-cotA-cotB-1=0

cotAcotB-cotA-cotB+1-2=0

cotA(cotB-1)-1(cotB-1)=2

(cotA-1)(cotB-1)=2

Question 16

यदि (If) $\tan \alpha-\tan \beta=x$ and $\cot \beta-\cot \alpha=y$
साबित करे(prove that)
$\cot (\alpha-\beta)=\frac{x+y}{x y}$
Sol :
$\tan \alpha-\tan \beta=x$

$\frac{\sin \alpha}{\cos \alpha}-\frac{\sin \beta}{\cos \beta}=x$

$\frac{\sin \alpha \cos \beta-\operatorname{cos} \alpha \sin \beta}{\cos \alpha \cos \beta}=x$

$\frac{\sin (\alpha-\beta)}{\cos \alpha \cos \beta}=x$

$\frac{\cos \alpha \cos \beta}{\sin (\alpha-\beta)}=\frac{1}{x}$..(i)

cotβ-cotα=y

$\frac{\cos \beta}{\sin \beta}-\frac{\cos \alpha}{\sin \alpha}=y$

$\frac{\sin \alpha \cos \beta-\cos \alpha \sin \beta}{\sin \beta \sin \alpha}=y$

$\frac{\sin (\alpha-\beta)}{\sin \alpha \sin \beta}=y$

$\frac{\sin \alpha \sin \beta}{\sin (\alpha-\beta)}=\frac{1}{y}$..(ii)

समीकरण (i) तथा (ii) को जोडने पर,
On adding (i) and (ii) ,

$\frac{\cos \alpha \cos \beta}{\sin(\alpha-\beta)}+\frac{\sin \alpha \sin \beta}{\sin (\alpha-\beta)}=\frac{1}{x}+\frac{1}{y}$

$\frac{\cos \alpha \cos \beta+\sin \alpha \sin \beta}{\sin (\alpha-\beta)}=\frac{y+x}{xy}$

$\frac{\cos (\alpha-\beta)}{\sin (\alpha-\beta)}=\frac{x+y}{x y}$

$\operatorname{cot}(\alpha-\beta)=\frac{x+y}{x y}$

Question 17

यदि एक समकोण तीन भागो 𝛼,ꞵ और 𝛾 मे विभाजित किया जाय तो साबित केर कि
(If a right angle be divided into three parts 𝛼,ꞵ and  𝛾 prove that)
$\cot \alpha=\frac{\tan \beta+\tan \gamma}{1-\tan \beta \tan \gamma}$
Sol :
α+β+ɣ=90°

β+ɣ=90°-α

taking both side tan

tan(β+ɣ)=tan(90°-α)

$\frac{\tan \beta+\tan \gamma}{1-\tan \beta \tan \gamma}=\cot \alpha$

Question 18

यदि(If) $2 \tan \beta+\cot \beta=\tan \alpha$तो साबित करे(show that)
$\cot \beta=2 \tan (\alpha-\beta)$
Sol :
$2 \tan \beta+\cot \beta=\tan \alpha$

$\tan \beta+\cot \beta=\tan \alpha-\tan \beta$

$\frac{\sin \beta}{\cos \beta}+\frac{\cos \beta}{\sin \beta}=\frac{\sin \alpha}{\cos \alpha}-\frac{\sin \beta}{\cos \beta}$

$\frac{\sin ^{2} \beta+\cos ^{2} \beta}{\cos \beta \sin \beta}=\frac{\sin \alpha \cos \beta-\cos \alpha \sin \beta}{\cos \alpha \cos \beta}$

$\frac{1}{\sin \beta}=\frac{\sin (\alpha-\beta)}{\cos \alpha}$..(i)

Again,(पुनः)

$2 \tan \beta+\operatorname{cot} \beta=\tan \alpha$

Adding both side $\cot \beta$

$2 \tan \beta+2 \cot \beta=\tan \alpha+\cot \beta$

$2(\tan \beta+\cot \beta=\tan \alpha+\cos \beta$

$2\left(\frac{\sin \beta}{\cos \beta}+\frac{\cos \beta}{\sin \beta}\right)=\frac{\sin \alpha}{\cos \alpha}+\frac{\cos \beta}{\sin \beta}$

$2\left(\frac{\sin ^{2} \beta+\cos ^{2} \beta}{\cos \beta \sin \beta}\right)=\frac{\sin \alpha \sin \beta+\cos \alpha \cos \beta}{\cos \alpha \sin \beta}$

$\frac{2}{\cos \beta}=\frac{\cos (\alpha-\beta)}{\cos \alpha}$..(ii)

समीकरण (i) मे (ii) से भाग देने पर
Dividing (i) by (ii)

$=\frac{\frac{1}{\sin \beta}}{\frac{2}{\cos \beta}}=\frac{\frac{\sin (\alpha-\beta)}{\cos \alpha}}{\frac{\cos (\alpha-B)}{\cos \alpha}}$

$\frac{\cos \beta}{2 \sin \beta}=\tan (\alpha-\beta)$

$\cot\beta=2 \tan (\alpha-\beta)$

Question 19

किसी ΔABC , मे C=90° , तो साबित करे कि
(If in any $\Delta \mathrm{ABC}, \mathrm{C}=90^{\circ}$ prove that)
$\operatorname{cosec}(A-B)=\frac{a^{2}+b^{2}}{a^{2}-b^{2}}$ (और)and $\sec (A-B)=\frac{c^{2}}{2 a b}
जहाँ A,B,C तीनो कोण है तथा a,b,c संगत भुजाएँ है ।
Sol :
Diagram

$\operatorname{cosec}(A-B)=\frac{1}{\sin (A-B)}$

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

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

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

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

$\operatorname{cosec}(A-B)=\frac{c^{2}}{a^{2}-b^{2}}$

$\text{cosec }(A-B)=\frac{a^{2}+b^{2}}{a^{2}-b^{2}}$

Question 19

किसी ΔABC , मे C=90° , तो साबित करे कि
(If in any $\Delta \mathrm{ABC}, \mathrm{C}=90^{\circ}$ prove that)
$\operatorname{cosec}(A-B)=\frac{a^{2}+b^{2}}{a^{2}-b^{2}}$ (और)and $\sec (A-B)=\frac{c^{2}}{2 a b}
जहाँ A,B,C तीनो कोण है तथा a,b,c संगत भुजाएँ है ।
Sol :

[$c^{2}=a^{2}+b^{2}$]

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

$=\frac{1}{\cos A \cos B+\sin A \sin B}$

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

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

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

$\sec (A-B)=\frac{C^{2}}{2 a b}$

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

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

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

$\sec (A-B)=\frac{C^{2}}{2 a b}$

Question 20

यदि(If) $\cot A=\sqrt{a c}$ , $\cot B=\sqrt{\frac{c}{a}}, \tan C=\sqrt{\frac{c}{a^{3}}}$ and $c=a^{2}+a+1$
साबित करे(prove that)
A+B=C
Sol :
$\tan A=\frac{1}{\sqrt{a{c}}}$ , $\tan B=\sqrt{\frac{a}{c}}$  , $\tan c=\sqrt{\frac{c}{a^{3}}}$ ,$c=a^{2}+a+1$

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

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

$\tan (A+B)=\frac{\frac{1+a}{\sqrt{a c}}}{\frac{\sqrt{a} \cdot c-\sqrt{a}}{\sqrt{a{c}} \cdot \sqrt{c}}}$

$\tan (A+B)=\frac{\sqrt{c}(1+a)}{\sqrt{a}(c-1)}$

$\left(\because C=a^{2}+a+1\right)$

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

$\tan (A+B)=\frac{\sqrt{c}(1+a)}{\sqrt{a} \cdot a(a+1)}$

$\tan (A+B)=\frac{\sqrt{c}}{\sqrt{a^{3}}}$

$\tan (A+B)=\sqrt{\frac{c}{a^{3}}}$

tan(A+B)=tan C

A+B=C

Question 21

यदि(If) $\frac{\tan (\mathrm{A}-\mathrm{B})}{\tan \mathrm{A}}+\frac{\sin ^{2} \mathrm{C}}{\sin ^{2} \mathrm{A}}=1$तो साबित करे कि(prove that)
tanA.tanB=$\tan ^{2} C$
Sol :
$\frac{\tan (\mathrm{A}-\mathrm{B})}{\tan \mathrm{A}}+\frac{\sin ^{2} \mathrm{C}}{\sin ^{2} \mathrm{A}}=1$

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

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

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

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

$\frac{\sin ^{2} C}{\sin ^{2} A}=\frac{\sin [A-(A-B)]}{\sin A \cos (A-D)}$

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

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

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

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

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

coaAcosB+sinAsinB-sinAsinB=$\cos^2 C$ cos(A-B)

$\cos ^{2} C \cos (A-B)=\cos A \cos B$..(ii)

समीकरण (i) मे (ii) से भाग देने पर
On dividing (i) by (ii)

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

$\tan ^{2}C$=tanA.tanB

Question 22

यदि(If) $\sin \alpha \sin \beta-\cos \alpha \cos \beta=1$साबित करे(show that)
$\tan \alpha+\tan \beta=0$
Sol :
$\sin \alpha \sin \beta-\cos \alpha \cos \beta=1$

$-1=\cos \alpha \cos \beta-\sin \alpha\sin \beta$

$-1=\cos (\alpha + \beta)$

$\sin (\alpha+\beta)=\sqrt{1-\cos ^{2}(\alpha+\beta)}$

$=\sqrt{1-(-1)^{2}}$

$=\sqrt{1-1}=0$

L.H.S
$\tan \alpha+\tan \beta=\frac{\sin \alpha}{\cos \alpha}+\frac{\sin \beta}{\cos \beta}$

$=\frac{\sin \alpha \cos \beta+\cos \alpha \sin \beta}{\cos \alpha \cos \beta}$

$=\frac{\sin (\alpha+B)}{\cos \alpha \cos \beta}$

$=\frac{0}{\cos \alpha \cos \beta}=0$

Question 23

यदि(If) sinθ=3sin(θ+2α) तो साबित करे(prove that)
tan(θ+α)+2tanα=0
Sol :
sinθ=3sin(θ+2α)

$\sin [(\theta+\alpha)-\alpha]=3 \sin ((\theta+\alpha)+\alpha]$

sin(θ+α).cosα-cos(θ+α)sinα=3[sin(θ+α)cosα+cos(θ+α)sinα]

sin(θ+α)cosα-cos(θ+α)sinα=3sin(θ+α)cosα+3cos(θ+α)sinα

sin(θ+α)cosα-3sin(θ+α)cosα=cos(θ+α)sinα+3cos(θ+α)sinα

-2sin(θ+α)cosα=4cos(θ+α)sinα

$\frac{\sin (\theta+\alpha)}{\cos (\theta+\alpha)}=\frac{4 \sin \alpha}{-2 \cos \alpha}$

tan(θ+α)=-2tanα

tan(θ+α)+2tanα=0

Question 24

यदि(If) mtan(θ-30°)=ntan(θ-120°)तो साबित करे(show that)
$\cos 2 \theta=\frac{m+n}{2(m-n)}$
Sol :
Componando and dividendo

If $\frac{a}{b}=\frac{c}{d}$ , then

$\frac{a+b}{a-b}=\frac{c+d}{c-d}$ or  $\frac{a-b}{a+b}=\frac{c-d}{c+d}$

Proof :
$\frac{a}{b}=\frac{c}{d}$

दोनो तरफ 1 जोडने पर
Adding both sides 1

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

$\frac{a+b}{b}=\frac{c+d}{d}$..(i)

(पुनः)
Again,$\frac{a}{b}=\frac{c}{d}$

दोनो तरफ 1 घटाने पर
Subtracting both sides 1

$\frac{a}{b}-1=\frac{c}{d}-1$

$\frac{a-b}{b}=\frac{c-d}{d}$..(ii)

समीकरण (i) मे (ii) से भाग देने पर
Dividing (i) by (ii)

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

$\frac{a+b}{a-b}=\frac{c+d}{c-d}$ or $\frac{a-b}{a+b}=\frac{c-d}{c+d}$

Question 24

यदि(If) $m \tan \left(\theta-30^{\circ}\right)=n \tan \left(\theta+120^{\circ}\right)$साबित करे(show that)
$\cos 2 \theta=\frac{m+n}{2(m-n)}$
Sol :
$m \tan \left(\theta-30^{\circ}\right)=n \tan \left(\theta+120^{\circ}\right)$

$\frac{m}{n}=\frac{\tan (\theta+120^{\circ})}{\tan (\theta-30^{\circ})}$

Componando and dividendo

$\frac{m+n}{m-n}=\frac{\tan (\theta+120^{\circ})+\tan (\theta-30^{\circ})}{\tan (\theta+120^{\circ})-\tan (\theta-30^{\circ})}$

$\frac{m+n}{m-n}=\frac{\frac{\sin (\theta+120^{\circ})}{\cos (\theta+120^{\circ})}+\frac{\tan (\theta-30^{\circ})}{\cos (\theta-30^{\circ})}}{\frac{\sin (\theta+120^{\circ})}{\cos (\theta+120^{\circ})}-\frac{\sin (\theta-30^{\circ})}{\cos (\theta-30^{\circ})}}$

$\frac{m+n}{m-n}=\dfrac{\frac{\sin (\theta+120^{\circ}) \cos (\theta-30^{\circ})+\cos (\theta+120^{\circ}) \tan (\theta-30^{\circ})}{\cos (\theta+120^{\circ}) \cos (\theta-30^{\circ})}}{\frac{\sin \left(\theta+120^{\circ}\right) \cos (\theta-30^{\circ})-\cos (\theta+120^{\circ}) \sin (\theta-30^{\circ})}{\cos (\theta+120^{\circ}) \cos (\theta-30^{\circ})}}$

$\frac{m+n}{m-n}=\frac{\sin [(\theta+120^{\circ})+(\theta-30^{\circ})]}{\sin [(\theta+120^{\circ})-(\theta-30^{\circ})]}$

$\frac{m+n}{m-n}=\frac{\sin [\theta+120^{\circ}+\theta-30^{\circ}]}{\sin [\theta+120^{\circ}-\theta+30^{\circ}]}$

$\frac{m+n}{m-n}=\frac{\sin (90^{\circ}+2 \theta)}{\sin 150^{\circ}}$

$\frac{m+n}{m-n}=\frac{\cos 2 \theta}{\sin (180^{\circ}-30^{\circ})}$

$\frac{m+n}{m-n}=\frac{\cos 2 \theta}{\sin 30^{\circ}}$

$\frac{m+n}{m-n}=\frac{\cos 2 \theta}{\frac{1}{2}}$

$\frac{m+n}{m-n}=2 \cos 2 \theta$

$\frac{m+n}{2(m-n)}=\cos 2 \theta$

Question 25

यदि(If) $\alpha+\beta=\theta$ (तथा)and $\tan \alpha: \tan \beta=x: y$(साबित करे)prove that
$\sin (\alpha-\beta)=\frac{x-y}{x+y} \sin \theta$
Sol :
$\frac{\tan \alpha}{\tan \beta}=\frac{x}{y}$

By Componendo and Dividendo

$\frac{\tan \alpha-\tan \beta}{\tan \alpha+\tan \beta}=\frac{x-y}{x+y}$

$\frac{\frac{\sin \alpha}{\cos \alpha}-\frac{\sin \beta}{\cos \beta}}{\frac{\sin \alpha}{\cos \alpha}+\frac{\sin \beta}{\cos \beta}}=\frac{x-y}{x+y}$

$\dfrac{\frac{\sin \alpha \cos \beta-\cos \alpha \sin \beta}{\cos \alpha \cos \beta}}{\frac{\sin \alpha \cos \beta+\cos \alpha \sin \beta}{\cos \alpha \cos \beta}}=\frac{x-y}{x+y}$

$\frac{\sin (\alpha-\beta)}{\sin (\alpha+\beta)}=\frac{x-y}{x+y}$

$\frac{\sin (\alpha-\beta)}{\sin \theta}=\frac{x-y}{x+y}$
$(\because \alpha+\beta=\theta)$

$\sin (\alpha-\beta)=\frac{x-y}{x+y} \sin \theta$

Question 26

साबित करे(Show that) sin100°-sin10° (धनातमक है)is positive
Sol :
sin100°-sin10°=sin(90°+10°)-sin10°

=cos10°-sin10°

$=\sqrt{2} \cdot \frac{1}{\sqrt{2}}(\cos 10^{\circ}-\sin 10^{\circ})$

$=\sqrt{2}\left(\frac{1}{\sqrt{2}} \cos 10^{\circ}-\frac{1}{\sqrt{2}} \sin 10^{\circ}\right)$

$=\sqrt{2} (\cos 45^{\circ} \cos 10^{\circ}-\sin 45^{\circ} \sin 10^{\circ})$

$=\sqrt{2} \cos (45+10)$

$=\sqrt{2} \cos 55^{\circ}>0$

Question 27

7cosθ+24sinθ का महतम और न्यूनतम मान निकाले।
Find the maximum and minimum values of 7cosθ+24sinθ
Sol :
(माना)Let 7=rsinx , 24=rcosx

दोनो को वर्ग करके जोड़ने पर
On squaring and then adding them

$7^{2}+24^{2}=r^{2} \sin ^{2} x+r^{2} \cos ^{2} x$

49+576=$r^{2}\left(\sin ^{2} x+\cos ^{2} x\right)$

$625=r^{2}$

$\sqrt{625}=r$

25=r

7cosθ+24sinθ=rsinx.cosθ+rcosx.sinθ

=r(sinx.cosθ+cosx.sinθ)

=rsin(x+θ)

=25sin(x+θ)

$\because-1 \leq \sin (x+\theta) \leq 1$

25 से गुणा करने पर,
Multiplying with 25

$-25 \leq 25\sin(x+8) \leq 25$

न्यूनतम मान
Minimum value=-25

महतम मान
Maximum value=25