frac{sec 8theta - 1}{sec 4theta - 1} is equal | vedclass

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

Question

(D) We have the expression $\frac{\sec 8	heta - 1}{\sec 4	heta - 1}$.Converting to cosine,we get $\frac{\frac{1}{\cos 8	heta} - 1}{\frac{1}{\cos 4\

Vedclass · Accepted Answer

$tan 8	heta \cot 2	heta$

Vedclass · Answer

(D) We have the expression $\frac{\sec 8	heta - 1}{\sec 4	heta - 1}$.Converting to cosine,we get $\frac{\frac{1}{\cos 8	heta} - 1}{\frac{1}{\cos 4	heta} - 1} = \frac{1 - \cos 8	heta}{\cos 8	heta} 	imes \frac{\cos 4	heta}{1 - \cos 4	heta}$.Using the identity $1 - \cos 2A = 2 \sin^2 A$,we have $1 - \cos 8	heta = 2 \sin^2 4	heta$ and $1 - \cos 4	heta = 2 \sin^2 2	heta$.Substituting these,we get $\frac{2 \sin^2 4	heta}{\cos 8	heta} 	imes \frac{\cos 4	heta}{2 \sin^2 2	heta}$.Expanding $\sin 4	heta = 2 \sin 2	heta \cos 2	heta$,we get $\frac{2(2 \sin 2	heta \cos 2	heta) \sin 4	heta \cos 4	heta}{\cos 8	heta \cdot 2 \sin^2 2	heta} = \frac{2 \sin 4	heta \cos 4	heta}{\cos 8	heta} 	imes \frac{2 \sin 2	heta \cos 2	heta}{2 \sin^2 2	heta}$.This simplifies to $\frac{\sin 8	heta}{\cos 8	heta} 	imes \frac{\cos 2	heta}{\sin 2	heta} = 	an 8	heta \cot 2	heta$.

$\frac{\sec 8\theta - 1}{\sec 4\theta - 1}$ is equal to

Similar Questions

If $\sin x + \cos x = \frac{1}{5},$ then $\tan 2x$ is

If $\alpha = \sin 36^{\circ}$,then $\alpha$ is a root of which of the following equations?

If the identity $\cos^4 \theta = a \cos 4\theta + b \cos 2\theta + c$ holds for some $a, b, c \in \mathbb{Q}$,then $(a, b, c) =$

Prove that $\frac{\sin 5x - 2\sin 3x + \sin x}{\cos 5x - \cos x} = \tan x$.

$\tan 20^\circ \cot 10^\circ \cot 50^\circ$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module