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

$	an 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 $540^{\circ} < \theta < 630^{\circ}$ and $\tan \theta = \frac{5}{12}$,then $\frac{\cos \frac{\theta}{2} - 5 \sin \frac{\theta}{2}}{\sqrt{-(12 \sec \theta + 5 \operatorname{cosec} \theta)}} = $

$\cos A \cos 2 A \cos 4 A \ldots \cos 2^{n-1} A$ equals

The value of $\cos \left(\frac{2 \pi}{7}\right) + \cos \left(\frac{4 \pi}{7}\right) + \cos \left(\frac{6 \pi}{7}\right)$ is equal to

If $\frac{5\pi}{2} < x < 3\pi$,then the value of the expression $\frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}}$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module