If cos theta = frac{-3}{5} and pi

Question

(B) Given,$	heta \in \left(\pi, \frac{3 \pi}{2}ight)$.Since $\pi < 	heta < \frac{3 \pi}{2}$,dividing by $2$ gives $\frac{\pi}{2} < \frac{	heta}{2

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(B) Given,$	heta \in \left(\pi, \frac{3 \pi}{2}ight)$.Since $\pi In this interval,$	an \left(\frac{	heta}{2}ight)$ is negative.Using the formula $\cos 	heta = \frac{1 - 	an^2(	heta/2)}{1 + 	an^2(	heta/2)}$,we have $\frac{1 - 	an^2(	heta/2)}{1 + 	an^2(	heta/2)} = \frac{-3}{5}$.Cross-multiplying gives $5 - 5 	an^2(	heta/2) = -3 - 3 	an^2(	heta/2)$.Rearranging terms,$8 = 2 	an^2(	heta/2)$,which implies $	an^2(	heta/2) = 4$.Thus,$	an(	heta/2) = \pm 2$.Since $\frac{	heta}{2}$ lies in the second quadrant,$	an(	heta/2)$ must be negative.Therefore,$	an(	heta/2) = -2$.

If $\cos \theta = \frac{-3}{5}$ and $\pi < \theta < \frac{3 \pi}{2}$,then $\tan \left(\frac{\theta}{2}\right) = $

Similar Questions

If $\sec x + \tan x = 2$,where $0 < x < \frac{\pi}{2}$,then find the value of $\sin \frac{x}{4}$.

If $90^{\circ} < A < 180^{\circ}$ and $\sin A = \frac{4}{5}$,then $\tan \frac{A}{2} =$

If $\theta = \frac{\pi}{9}$,then $1 + 27 \tan^2 \theta - 33 \tan^4 \theta + \tan^6 \theta = $

$\frac{\cos A}{1 - \sin A} = $

$\frac{\sec 8A - 1}{\sec 4A - 1} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module