If sec(theta+alpha),sectheta,and sec(theta-al | vedclass

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

Question

(D) Given that $\sec(	heta+\alpha)$,$\sec	heta$,and $\sec(	heta-\alpha)$ are in arithmetic progression $(AP)$.Therefore,$2 \sec	heta = \sec(	heta

Vedclass · Accepted Answer

$\pm \sqrt{2}$

Vedclass · Answer

(D) Given that $\sec(	heta+\alpha)$,$\sec	heta$,and $\sec(	heta-\alpha)$ are in arithmetic progression $(AP)$.Therefore,$2 \sec	heta = \sec(	heta+\alpha) + \sec(	heta-\alpha)$.Using the identity $\sec x = \frac{1}{\cos x}$,we have:$\frac{2}{\cos	heta} = \frac{1}{\cos(	heta+\alpha)} + \frac{1}{\cos(	heta-\alpha)}$.$\frac{2}{\cos	heta} = \frac{\cos(	heta-\alpha) + \cos(	heta+\alpha)}{\cos(	heta+\alpha)\cos(	heta-\alpha)}$.Using the sum-to-product formula $\cos(A-B) + \cos(A+B) = 2\cos A \cos B$:$\frac{2}{\cos	heta} = \frac{2\cos	heta \cos\alpha}{\cos^2	heta - \sin^2\alpha}$.$\frac{1}{\cos	heta} = \frac{\cos	heta \cos\alpha}{\cos^2	heta - \sin^2\alpha}$.$\cos^2	heta - \sin^2\alpha = \cos^2	heta \cos\alpha$.$\cos^2	heta(1 - \cos\alpha) = \sin^2\alpha$.$\cos^2	heta(2\sin^2\frac{\alpha}{2}) = (2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2})^2$.$\cos^2	heta(2\sin^2\frac{\alpha}{2}) = 4\sin^2\frac{\alpha}{2}\cos^2\frac{\alpha}{2}$.Assuming $\sin\frac{\alpha}{2} 
eq 0$,we get $\cos^2	heta = 2\cos^2\frac{\alpha}{2}$.Taking the square root,$\cos	heta = \pm \sqrt{2} \cos\frac{\alpha}{2}$.Therefore,$\cos	heta \cdot \sec\frac{\alpha}{2} = \pm \sqrt{2}$.

If $\sec(\theta+\alpha)$,$\sec\theta$,and $\sec(\theta-\alpha)$ are in arithmetic progression,then the value of $\cos\theta \cdot \sec\frac{\alpha}{2}$ is:

Similar Questions

It is known that $\sin \beta = \frac{4}{5}$ and $0 < \beta < \pi$. Then the value of $\frac{\sqrt{3} \sin(\alpha + \beta) - \frac{2}{\cos(\pi/6)} \cos(\alpha + \beta)}{\sin \alpha}$ is:

If $x=\log _e\left[\cot \left(\frac{\pi}{4}+\theta\right)\right]$ and $\theta \in\left(\frac{-\pi}{4}, \frac{\pi}{4}\right)$,then consider the following statements:
$I$. $\cosh x=\sec 2 \theta$
$II$. $\sinh x=-\tan 2 \theta$

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

$\tan ^2 \frac{\pi}{16}+\tan ^2 \frac{2 \pi}{16}+\tan ^2 \frac{3 \pi}{16}+\tan ^2 \frac{4 \pi}{16}+\tan ^2 \frac{5 \pi}{16}+\tan ^2 \frac{6 \pi}{16}+\tan ^2 \frac{7 \pi}{16} = ?$

Evaluate: $\cos^2 76^\circ + \cos^2 16^\circ - \cos 76^\circ \cos 16^\circ$

Vedclass Test Series

Exam Paper Generator

Online Exam Module