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) = $

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