Suppose theta and phi neq 0 are such that sec | vedclass

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

Question

(A) Since $\sec(	heta - \phi)$,$\sec 	heta$,and $\sec(	heta + \phi)$ are in $A.P.$,we have:$2 \sec 	heta = \sec(	heta - \phi) + \sec(	heta + \ph

Vedclass · Accepted Answer

$\pm \sqrt{2}$

Vedclass · Answer

(A) Since $\sec(	heta - \phi)$,$\sec 	heta$,and $\sec(	heta + \phi)$ are in $A.P.$,we have:$2 \sec 	heta = \sec(	heta - \phi) + \sec(	heta + \phi)$$\frac{2}{\cos 	heta} = \frac{\cos(	heta + \phi) + \cos(	heta - \phi)}{\cos(	heta - \phi) \cos(	heta + \phi)}$Using the identity $\cos(A+B) + \cos(A-B) = 2 \cos A \cos B$ and $\cos(A-B) \cos(A+B) = \cos^2 A - \sin^2 B$:$\frac{2}{\cos 	heta} = \frac{2 \cos 	heta \cos \phi}{\cos^2 	heta - \sin^2 \phi}$$\cos^2 	heta - \sin^2 \phi = \cos^2 	heta \cos \phi$$\cos^2 	heta (1 - \cos \phi) = \sin^2 \phi$$\cos^2 	heta (1 - \cos \phi) = 1 - \cos^2 \phi = (1 - \cos \phi)(1 + \cos \phi)$Since $\phi 
eq 0$,$1 - \cos \phi 
eq 0$,so:$\cos^2 	heta = 1 + \cos \phi = 2 \cos^2(\frac{\phi}{2})$$\cos 	heta = \pm \sqrt{2} \cos(\frac{\phi}{2})$Given $\cos 	heta = k \cos(\frac{\phi}{2})$,we find $k = \pm \sqrt{2}$.

Suppose $\theta$ and $\phi \neq 0$ are such that $\sec(\theta + \phi)$,$\sec \theta$,and $\sec(\theta - \phi)$ are in $A.P.$ If $\cos \theta = k \cos(\frac{\phi}{2})$ for some $k$,then $k$ is equal to

Similar Questions

If $A$ lies in the third quadrant and $3 \tan A - 4 = 0,$ then $5 \sin 2A + 3 \sin A + 4 \cos A = $

If $x + y = 3 - \cos 4\theta$ and $x - y = 4 \sin 2\theta$,then:

$3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4(\sin ^{6} x+\cos ^{6} x)$ is equal to

If $\cos x + \cos y = \frac{2}{3}$ and $\sin x - \sin y = \frac{3}{4}$,then find the value of $\sin(x - y) + \cos(x - y)$.

If $3 \sin \alpha = 5 \sin \beta$,then $\tan \left(\frac{\alpha + \beta}{2}\right) \div \tan \left(\frac{\alpha - \beta}{2}\right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module