KC Sinha Mathematics Solution Class 11 Chapter 6 त्रिकोणमितीय फलन (Trigonometric function) Exercise 6.4

Exercise 6.4

Question 1

(i) Prove that :

$\frac{2 \sin \theta-\sin 2 \theta}{2 \sin \theta+\sin 2 \theta}=\tan ^{2} \frac{\theta}{2}$
Sol :
L.H.S

$=\frac{2 \sin \theta-\sin 2 \theta}{2 \sin \theta+\sin 2 \theta}$

$=\frac{2 \sin \theta-2 \sin \theta \cos \theta}{2 \sin \theta+2 \sin \theta \cos \theta}$

$=\frac{2 \sin \theta(1-\cos \theta)}{2 \sin \theta(1+\cos \theta)}$

$=\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}$

$=\tan ^{2} \frac{\theta}{2}$

(ii)

$\sin 3 x+\sin 2 x-\sin x=4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2}$
Sol :
L.H.S

=sin3x+sin2x-sinx

$=2 \cos \frac{3 x+x}{2} \sin \frac{3 x-x}{2}+\sin 2 x$

=2cos2xsinx+2sinxcosx

$\because \cos C+\cos D=2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$

=2sinx(cos2x+cosx)

$=2 \sin x \cdot 2 \cos \frac{2 x+x}{2} \cos \frac{2 x-x}{2}$

$=4 \sin x \cos \frac{3 x}{2} \cos \frac{x}{2}$

Question 2

Prove that

$\cot \frac{\theta}{2}-\tan \frac{\theta}{2}=2 \cot \theta$
Sol :
L.H.S

$\cot \frac{\theta}{2}-\tan \frac{\theta}{2}$

$=\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}-\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}$

$=\frac{\cos ^{2} \frac{\theta}{2}-\sin ^{2} \frac{\theta}{2}}{\sin \frac{\theta}{2} \cos \frac{\theta}{2}}$

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

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

=2cotθ

Question 3

Prove that :

$\frac{1+\sin \theta}{1-\sin \theta}=\tan ^{2}\left(\frac{\pi}{4}+\frac{\theta}{2}\right)$
Sol :
L.H.S

$\frac{1+\sin \theta}{1-\sin \theta}=\frac{\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)^{2}}{\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)^{2}}$

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

$=\left[\dfrac{\frac{\cos \frac{\theta}{2}}{\cos \frac{\theta}{2}}+\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}}{\frac{\cos \frac{\theta}{2}}{\operatorname{cos} \frac{\theta}{2}}-\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}}\right]^2$

$=\left[\frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}}\right]^{2}$

$=\left[\frac{\tan \frac{\pi}{7}+\tan \frac{\theta}{2}}{1-\tan \frac{\pi}{4} \cdot \tan \frac{\theta}{2}}\right]^{2}$

=\tan ^{2}\left(\frac{\pi}{4}+\frac{\theta}{2}\right)

Question 4

$\sec \theta+\tan \theta=\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)$
Sol :
L.H.S
=secθ+tanθ

$=\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}$

$=\frac{1+\sin \theta}{\cos \theta}$

$=\frac{\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)^{2}}{\cos ^{2} \frac{\theta}{2}-\sin ^{2} \frac{\theta}{2}}$

$=\frac{\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)^{2}}{\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)\left(\cos \frac{\theta}{2}+\operatorname{\sin} \frac{\theta}{2}\right)}$

$=\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}$

[Dividing by

$\cos \frac{\theta}{2}$ ]

$=\frac{\frac{\cos \frac{\theta}{2}}{\cos \frac{\theta}{2}}+\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}}{\frac{\cos \frac{\theta}{2}}{\cos \frac{\theta}{2}}-\frac{\operatorname{sin}^{2} \frac{\theta}{2}}{\cos \frac{\theta}{2}}}$

$=\frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}}$

$=\frac{\tan \frac{\pi}{4}+\tan \frac{\theta}{2}}{1-\tan \frac{\pi}{4} \tan \frac{\theta}{2}}$

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

Question 5

Prove that :

$\frac{\sin \alpha+\sin \beta-\sin (\alpha+\beta)}{\sin \alpha+\sin \beta+\sin (\alpha+\beta)}=\tan \frac{\alpha}{2} \tan \frac{\beta}{2}$
Sol :
L.H.S

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

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

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

$=\frac{2\sin \frac{\alpha+\beta}{2}\left[\cos \frac{\alpha-\beta}{2}\cos \frac{\alpha+\beta}{2}\right]}{2 \sin \frac{\alpha+\beta}{2}\left[\cos \frac{\alpha-\beta}{2}+\cos \frac{\alpha+\beta}{2}\right]}$

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

$=\tan \frac{\alpha}{2} \tan \frac{\beta}{2}$

Question 6

Prove that:

$\operatorname{cosec}\left(\frac{\pi}{4}+\frac{\theta}{2}\right) \operatorname{cosec}\left(\frac{\pi}{4}-\frac{\theta}{2}\right)=2 \sec \theta$
Sol :
L.H.S

$\operatorname{cosec}\left(\frac{\pi}{4}+\frac{\theta}{2}\right) \operatorname{cosec}\left(\frac{\pi}{4}-\frac{\theta}{2}\right)$

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

[∵ sin(A+B)sin(A-B)=sin²A-sin²B]

$=\frac{1}{\sin ^{2} \frac{\pi}{4}-\sin ^{2} \frac{\theta}{2}}$

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

$=\frac{1}{\frac{1}{2}-\frac{\sin ^{2} \frac{\theta}{2}}{1}}$

$=\frac{1}{\frac{1-2 \sin ^{2} \frac{\theta}{2}}{2}}$

$=\frac{2}{1-2 \sin ^{2} \frac{\theta}{2}}$

$=\frac{2}{\cos \theta}$

=2secθ

Question 8

Prove that:

$(\cos A-\cos B)^{2}+(\sin A-\sin B)^{2}=4 \sin ^{2} \frac{A-B}{2}$
Sol :
L.H.S

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

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

$=4 \sin ^{2} \frac{A+B}{2} \sin ^{2} \frac{A-B}{2}++\cos ^{2} \frac{A+B}{2} \sin ^{2} \frac{A-B}{2}$

$=4 \sin ^{2} \frac{A-B}{2}\left[\sin ^{2} \frac{A+B}{2}+\cos ^{2} \frac{A+B}{2}\right]$

$=4 \sin ^{2} \frac{A-B}{2}$

Question 9

Prove that :

$\cos ^{2} \frac{\pi}{8}+\cos ^{2} \frac{3 \pi}{8}+\cos ^{2} \frac{5 \pi}{8}+\cos ^{2} \frac{7 \pi}{8}=2$
Sol :
L.H.S

$\cos ^{2} \frac{\pi}{8}+\cos ^{2} \frac{3 \pi}{8}+\cos ^{2} \frac{5 \pi}{8}+\cos ^{2} \frac{7 \pi}{8}$

$=\cos ^{2} \frac{\pi}{8}+\cos ^{2} \frac{3 \pi}{8}+\left[\cos \left(\frac{\pi}{2}+\frac{\pi}{8}\right)\right]^{2}+\left[\cos \left(\frac{\pi}{2}+\frac{3 \pi}{3}\right)\right]^{2}$

$=\cos ^{2} \frac{\pi}{8}+\cos ^{2} \frac{3 \pi}{8}+\left[-\sin \frac{\pi}{8}\right]^{2}+\left[-\sin \frac{3 \pi}{8}\right]^{2}$

$=\cos \frac{2 \pi}{8}+\cos ^{2} \frac{3 \pi}{8}+\sin ^{2} \frac{\pi}{8}+\sin ^{2} \frac{3}{8}$

=1+1=2

Question 11

Prove that :

$\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)=\frac{1}{8}$
Sol :
L.H.S

$\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{3}\right)$

$=\left(1+\cos \frac{{\pi}}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left[1+\cos \left(\pi-\frac{3\pi}{8}\right)\right]\left[1+\cos \left(\pi-\frac{\pi}{8}\right)\right.$

$=\left(1+\cos \frac{\pi}{3}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{\pi}{8}\right)$

$=\left[1^{2}-\cos ^{2} \frac{\pi}{8}\right]\left[1^{2}-\cos ^{2} \frac{3 \pi}{3}\right]$

$=\left[1-\cos ^{2} \frac{\pi}{8}\right]\left[1-\cos ^{2}\left(\frac{\pi}{2}-\frac{\pi}{8}\right)\right]$

$=\left[1-\cos ^{2} \frac{\pi}{8}\right]\left[1-\sin ^{2} \frac{\pi}{8}\right]$

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

$=\left[1-\frac{2+\sqrt{2}}{4}\right]\left[1-\frac{2-\sqrt{2}}{4}\right]$

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

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

$=\frac{2^{2}-(\sqrt{2})^{2}}{16}$

$=\frac{4-2}{16}=\frac{2}{16}$

Question 12

Prove that:

$\cot 142 \frac{1}{2}^{\circ}=\sqrt{2}+\sqrt{3}-2-\sqrt{6}$
Sol :
L.H.S

$\cot 142 \frac{1}{2}^{\circ}=\cot \left(180^{\circ}-37 \frac{1}{2}^{\circ}\right)$

$=-\cot 37\frac{1}{2}$

$\cot \frac{A}{2}=\frac{1+\cos A}{\sin A}$

A=75 ,

$\cot \frac{75}{2}^{\circ}=\frac{1+\cos 75^{\circ}}{\sin 75^{\circ}}$

$\cot 37 \frac{1}{2}^{\circ}=\frac{1+\sin 15^{\circ}}{\cos 15^{\circ}}$

$\cot 37 \frac{1}{2}^{\circ}=\frac{1+\frac{\sqrt{3}-1}{2 \sqrt{2}}}{\frac{\sqrt{3}+1}{2 \sqrt{2}}}$

$\cot 37 \frac{1}{2}^{\circ}=\frac{\frac{2 \sqrt2+\sqrt{3}-1}{2 \sqrt{2}}}{\frac{\sqrt{3}+1}{2 \sqrt{2}}}$

$\cot 37 \frac{1}{2}^{\circ}=\frac{2 \sqrt{2}+\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt 3-1}{\sqrt{3}-1}$

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

$=\frac{2 \sqrt{6}-2 \sqrt{3}+4-2 \sqrt{2}}{2}$

$=\frac{2\left(\sqrt6-\sqrt{3}+2-\sqrt{2}\right)}{2}$

$\cot 37 \frac{1}{2}^{\circ}=\sqrt{6}-\sqrt{3}+2-\sqrt{2}$

$\cot 1+2 \frac{1}{2}^{\circ}=-\cot 37 \frac{1}{2}^{\circ}$

=-(√6-√3+2-√2)
=-√6+√3-2+√2
=√3+√2-2-√6

Question 13

Prove that:

$\sin ^{2} 48^{\circ}-\cos ^{2} 12^{\circ}=\frac{\sqrt{5}+1}{8}$
Sol :
L.H.S

$\sin ^{2} 48^{\circ}-\cos ^{2} 12^{\circ}$

$=\sin ^{2} 48^{\circ}-\sin ^{2} 78^{\circ}$

$=\sin \left(48^{\circ}+78^{\circ}\right) \cdot \operatorname{sin}\left(48^{\circ}-78^{\circ}\right)$

$=\sin 126^{\circ}\sin(-30^{\circ})$

=-sin126°sin30°

=-sin(90°+36°)sin30°

=-cos36°sin30°

$=-\frac{\sqrt{5}+1}{4} \times \frac{1}{2}$

$=-\left(\frac{(\sqrt5+1)}{8}\right)$

Question 22

$\tan \frac{\theta}{2}=\sqrt{\frac{1-e}{1+e} \tan \frac{\phi}{2}}$ prove that

$\cos \phi=\frac{\cos \theta-e}{1-e \cos \theta}$
Sol :

$\tan \frac{\theta}{2}=\sqrt{\frac{1-e}{1+e}} \tan \frac{\phi}{2}$

Squaring both sides

$\tan ^{2} \frac{\theta}{2}=\frac{1-e}{1+e} \tan ^{2} \frac{\phi}{2}$

$\frac{1+e}{1-e} \tan ^{2} \frac{\theta}{2}=\tan ^{2} \frac{\phi}{2}$

R.H.S

$\frac{\cos \theta-e}{1-e \cos \theta}$

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

$=\frac{\frac{1-\tan ^{2} \frac{\theta}{2}-e-e \tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}}{\frac{1+\tan ^{2} \frac{\theta}{2}-e+e \tan ^{2} \frac{\theta}{2}}{1+\tan ^{2}{\frac{\theta}{2}}}}$

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

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

$=\frac{1-\tan ^{2} \frac{\phi}{2}}{1+\tan ^{2} \frac{\phi}{2}}$

=cosɸ

Question 23

If sin𝛼+sinβ=a and cos𝛼+cosβ=b prove that:

(i)

$\sin (\alpha+\beta)=\frac{2 a b}{a^{2}+b^{2}}$
Sol :
sin𝛼+sinβ=a

$2 \sin \frac{\alpha+\beta}{2} \cos \alpha-\frac{B}{2}=a$ ..(i)

Similarly:

cos𝛼+cosβ=b

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

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

$\tan \frac{\alpha+\beta}{2}=\frac{a}{b}$

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

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

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

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

$\sin (\alpha+\beta)=\frac{2 a b}{a^{2}+b^{2}}$

(ii)

$\cos (\alpha-\beta)=\frac{1}{2}\left(a^{2}+b^{2}-2\right)$
Sol :
sin𝛼+sinβ=a..(i)

and

cos𝛼+cosβ=b..(ii)

On squaring and adding

$(\sin \alpha+\sin \beta)^{2}+(\cos \alpha+\cos \beta)^{2}=a^{2}+b^{2}$

$\sin^{2} \alpha+\sin ^{2} \beta+2 \sin \alpha \sin \beta+\cos ^{2} \alpha+\cos^{2} \beta+2 \cos \alpha \cos \beta=a^{2}+b^{2}$

$1+1+2(\cos \alpha \cos \beta+\sin \alpha \sin \beta)=a^{2}+b^{2}$

$2 \cos (\alpha-\beta)=a^{2}+b^{2}-2$

$\cos (\alpha-\beta)=\frac{1}{2}\left(a^{2}+b^{2}-2\right)$

Question 24

(i)

$\tan (\alpha+\beta)=\frac{2 a b}{a^{2}+b^{2}}$
Sol :
acos𝛼+bsin𝛼=C..(i)

acosβ+bsinβ=C..(ii)

Subtracting (i) from (ii)

a(cos𝛼-cosβ)+b(sin𝛼-sinβ)=0

$a\left(-2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}\right)+b \left( 2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}\right)=0$

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

$\frac{b}{a}=\tan \frac{\alpha+\beta}{2}$

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

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

$\tan (\alpha+\beta)=\frac{\frac{2 b}{a}}{\frac{a^{2}-b^{2}}{a^{2}}}$

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

(ii)

$\cos (\alpha+\beta)=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$
Sol :

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

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

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

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

Question 26

$2 \tan \frac{\alpha}{2}=\tan \frac{\beta}{2}$ Prove that

$\cos \alpha=\frac{3+5 \cos \beta}{5+3 \cos \beta}$
Sol :
R.H.S

$=\frac{3+5 \cos \beta}{5+3 \cos \beta}$

$=\frac{3+5\left(\frac{1-\tan ^{2} \frac{\beta}{2}}{1+\tan ^{2} \frac{\beta}{2}}\right)}{5+3\left(\frac{1-\tan ^{2} \frac{\beta}{2}}{1+\tan ^{2} \frac{\beta}{2}}\right)}$

$=\frac{\frac{3+3 \tan ^{2} \frac{\beta}{2}+5-5 \tan ^{2} \frac{\beta}{2}}{2}}{\left.\frac{5+5 \tan ^{2} \frac{\beta}{2}+3-3\tan ^{2} \frac{\beta}{2}}{2+\tan ^{2} \frac{\beta}{2}}\right.}$

$=\frac{8-2 \tan ^{2} \frac{\beta}{2}}{8+2 \tan ^{2} \frac{\beta}{2}}$

$=\frac{8-2\left(2 \tan \frac{\alpha}{2}\right)^{2}}{8+2\left(2 \tan \frac{\alpha}{2}\right)^{2}}$

$=\frac{8-8 \tan ^{2} \frac{\alpha}{2}}{8+8 \tan ^{2} \frac{\alpha}{2}}$

$=\frac{8\left(1-\tan ^{2} \frac{\alpha}{2}\right)}{8\left(1+\tan ^{2} \frac{\alpha}{2}\right)}$

=cos𝛼

Question 27

$\sin \alpha=\frac{4}{5}$ and

$\cos \beta=\frac{5}{13}$ prove that

$\cos \frac{\alpha-\beta}{2}=\frac{8}{\sqrt{65}}$
Sol :

$\sin \alpha=\frac{4}{5} \Rightarrow \cos \alpha=\sqrt{1-\sin ^{2}} \alpha$

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

$=\sqrt{1-\frac{16}{25}}$

$=\sqrt{\frac{25-16}{25}}$

$=\sqrt{\frac{9}{25}}=\frac{3}{5}$

$\cos \alpha=\frac{3}{5}$

$\cos \beta=\frac{5}{13}, \Rightarrow \sin \beta=\sqrt{1-\cos ^{2} \beta}=\sqrt{1-\left(\frac{5}{13}\right)^{2}}$

$=\frac{12}{13}$

cos(𝛼-β)=cos𝛼cosβ+sin𝛼sinβ

$=\frac{3}{5} \times \frac{5}{13}+\frac{4}{5} \times \frac{12}{13}$

$=\frac{15}{65}+\frac{48}{65}=\frac{63}{65}$

$\cos(\alpha-\beta)=\frac{63}{65}$

$2 \cos ^{2}\left(\frac{\alpha-\beta}{2}\right)-1=\frac{63}{65}$

$2 \cos ^{2}\left(\frac{\alpha-\beta}{2}\right)=\frac{63}{65}+1$

$2 \cos ^{2} \frac{\alpha-\beta}{2}=\frac{63+65}{65}$

$2 \cos ^{2} \frac{\alpha-\beta}{2}=\frac{128}{65}$

$\cos ^{2} \frac{\alpha-\beta}{2}=\frac{128}{65 \times 2}$

$\cos \frac{\alpha-\beta}{2}=\sqrt{\frac{64}{65}}=\frac{8}{\sqrt{65}}$

Question 28

If sec(ɸ+𝛼)+sec(ɸ-𝛼)=2secɸ prove that

$\cos \phi=\pm \sqrt{2 \cos \frac{\alpha}{2}}$
Sol :
sec(ɸ+𝛼)+sec(ɸ-𝛼)=2secɸ

$\frac{1}{\cos (\phi+\alpha)}+\frac{1}{\cos (\phi-\alpha)}=\frac{2}{\cos \phi}$

$\frac{\cos(\phi-\alpha)+\cos (\phi+\alpha)}{\cos (\phi+\alpha) \cos (\phi-\alpha)}=\frac{2}{\cos \phi}$

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

$\cos^{2} \phi \cos \alpha=\cos ^{2} \phi-\sin ^{2} \alpha$

$\sin ^{2} \alpha=\cos ^{2} \phi-\cos ^{2} \phi \cos \alpha$

$\sin ^{2} x=\cos ^{2} \phi(1-\cos \alpha)$

$\left(2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}\right)^{2}=\cos ^{2} \phi \cdot 2 \sin ^{2} \frac{\alpha}{2}$

$4 \sin ^{2} \frac{\alpha}{2} \cos ^{2} \frac{\alpha}{2}=\cos ^{2} \phi \cdot 2 \sin ^{2} \frac{\alpha}{2}$

$2 \cos ^{2} \frac{\alpha}{2}=\cos ^{2} \phi$

$\pm \sqrt{2 \cos ^{2} \frac{\alpha}{2}}=\cos \phi$

$\pm \sqrt{2} \cos \frac{\alpha}{2}=\cos \phi$