If tan^2 theta = 2tan^2 phi + 1,then cos 2the | vedclass

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

Question

(B) Given: $	an^2 	heta = 2	an^2 \phi + 1$ Adding $1$ to both sides: $1 + 	an^2 	heta = 2	an^2 \phi + 2$ $\Rightarrow \sec^2 	heta = 2(1 + 	an

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Given: $	an^2 	heta = 2	an^2 \phi + 1$ Adding $1$ to both sides: $1 + 	an^2 	heta = 2	an^2 \phi + 2$ $\Rightarrow \sec^2 	heta = 2(1 + 	an^2 \phi)$ $\Rightarrow \sec^2 	heta = 2\sec^2 \phi$ Taking the reciprocal: $\cos^2 	heta = \frac{1}{2} \cos^2 \phi$ $\Rightarrow 2\cos^2 	heta = \cos^2 \phi$ Using the identity $\cos 2	heta = 2\cos^2 	heta - 1$,we have $2\cos^2 	heta = 1 + \cos 2	heta$ Also,$\cos^2 \phi = 1 - \sin^2 \phi$ Substituting these into the equation $2\cos^2 	heta = \cos^2 \phi$: $1 + \cos 2	heta = 1 - \sin^2 \phi$ $\Rightarrow \cos 2	heta + \sin^2 \phi = 0$.

If $\tan^2 \theta = 2\tan^2 \phi + 1$,then $\cos 2\theta + \sin^2 \phi$ equals

Similar Questions

$\tan \frac{\pi}{5} + 2 \tan \frac{2 \pi}{5} + 4 \cot \frac{4 \pi}{5} = $

If $a = \sin \frac{\pi}{18} \sin \frac{5\pi}{18} \sin \frac{7\pi}{18}$ and $x$ is the solution of the equations $y = 2[x] + 2$ and $y = 3[x - 2]$,where $[x]$ denotes the greatest integer function of $x$,then $a$ is equal to:

The value of $\tan 20^{\circ} \tan 80^{\circ} \cot 50^{\circ} = $

If $\cos \theta = \frac{\cos \alpha - \cos \beta}{1 - \cos \alpha \cos \beta}$,then one of the values of $\tan \frac{\theta}{2}$ is

If $\operatorname{Sinh}^{-1} x = \log 3$ and $\operatorname{Cosh}^{-1} y = \log \frac{3}{2}$,then $\operatorname{Tanh}^{-1}(x-y) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module