Let f_n(x) = tan frac{x}{2}(1 + sec x)(1 + se | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We know that $(1 + \sec 	heta) = \frac{1 + \cos 	heta}{\cos 	heta} = \frac{2 \cos^2 \frac{	heta}{2}}{\cos 	heta}$.Also, $	an \frac{	heta}{2

Vedclass · Accepted Answer

$f_2\left(\frac{\pi}{16}\right) = 1$

Vedclass · Answer

(D) We know that $(1 + \sec 	heta) = \frac{1 + \cos 	heta}{\cos 	heta} = \frac{2 \cos^2 \frac{	heta}{2}}{\cos 	heta}$.Also, $	an \frac{	heta}{2} (1 + \sec 	heta) = \frac{\sin \frac{	heta}{2}}{\cos \frac{	heta}{2}} \cdot \frac{2 \cos^2 \frac{	heta}{2}}{\cos 	heta} = \frac{2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}}{\cos 	heta} = \frac{\sin 	heta}{\cos 	heta} = 	an 	heta$.Applying this repeatedly, $f_n(x) = 	an \frac{x}{2} (1 + \sec x) (1 + \sec 2x) \dots (1 + \sec 2^n x) = 	an 2^n x$.For $x = \frac{\pi}{16}$:$f_2\left(\frac{\pi}{16}ight) = 	an \left(2^2 \cdot \frac{\pi}{16}ight) = 	an \left(4 \cdot \frac{\pi}{16}ight) = 	an \frac{\pi}{4} = 1$.Thus, option $D$ is correct.

Let $f_n(x) = \tan \frac{x}{2}(1 + \sec x)(1 + \sec 2x) \dots (1 + \sec 2^n x)$, then

Similar Questions

$\tan \alpha + 2 \tan 2 \alpha + 4 \tan 4 \alpha + 8 \cot 8 \alpha = $

$2\cos x - \cos 3x - \cos 5x = $

If $\tan x = \frac{3}{4}$ and $\pi < x < \frac{3 \pi}{2}$,then $\cos \frac{x}{2} = . . . . . .$

$\tan A + 2 \tan 2A + 4 \tan 4A + 8 \cot 8A = $

$\cot \theta - \tan \theta - 2 \tan 2 \theta - 4 \tan 4 \theta = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module