frac{sec 8A - 1}{sec 4A - 1} = | vedclass

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

Question

(B) આપેલ પદાવલિ: $\frac{\sec 8A - 1}{\sec 4A - 1}$$= \frac{\frac{1}{\cos 8A} - 1}{\frac{1}{\cos 4A} - 1} = \frac{1 - \cos 8A}{\cos 8A} 	imes \frac{\c

Vedclass · Accepted Answer

$\frac{	an 8A}{	an 2A}$

Vedclass · Answer

(B) આપેલ પદાવલિ: $\frac{\sec 8A - 1}{\sec 4A - 1}$$= \frac{\frac{1}{\cos 8A} - 1}{\frac{1}{\cos 4A} - 1} = \frac{1 - \cos 8A}{\cos 8A} 	imes \frac{\cos 4A}{1 - \cos 4A}$નિત્યસમ $1 - \cos 	heta = 2 \sin^2(\frac{	heta}{2})$ નો ઉપયોગ કરતા:$= \frac{2 \sin^2 4A}{\cos 8A} 	imes \frac{\cos 4A}{2 \sin^2 2A}$$= \frac{(2 \sin 4A \cos 4A) \sin 4A}{\cos 8A 	imes 2 \sin^2 2A}$કારણ કે $2 \sin 4A \cos 4A = \sin 8A$:$= \frac{\sin 8A}{\cos 8A} 	imes \frac{\sin 4A}{2 \sin^2 2A} = 	an 8A 	imes \frac{2 \sin 2A \cos 2A}{2 \sin^2 2A}$$= 	an 8A 	imes \frac{\cos 2A}{\sin 2A} = 	an 8A 	imes \cot 2A = \frac{	an 8A}{	an 2A}$

$\frac{\sec 8A - 1}{\sec 4A - 1} = $

Similar Questions

જો $\tan A = \frac{1}{2}$ અને $\tan B = \frac{1}{3}$ હોય,તો $\tan (A + 2B)$ ની કિંમત શોધો.

$\cos \frac{\pi }{7} \cos \frac{2\pi }{7} \cos \frac{4\pi }{7} = $

$\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{3\pi}{7} =$

$\tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ = $

$\cot x - \tan x = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module