જો tan theta = frac{{ - 4}}{3}, તો&nbsp | vedclass

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(b) Since ${m{cose}}{{m{c}}^2}	heta = 1 + {\cot ^2}	heta = 1 + \frac{9}{{16}} = \frac{{25}}{{16}}$

$\left( \because   {	an 	heta&

Vedclass · Accepted Answer

$-4/5$ અથવા $4/5$

Vedclass · Answer

(b) Since ${m{cose}}{{m{c}}^2}	heta = 1 + {\cot ^2}	heta = 1 + \frac{9}{{16}} = \frac{{25}}{{16}}$

$\left( \because   {	an 	heta  =  - \frac{4}{3}} ight)$

${\sin ^2}	heta = \frac{1}{{{m{cose}}{{m{c}}^2}	heta }} = \frac{{16}}{{25}} $

$\Rightarrow \sin 	heta = \pm \frac{4}{5},$

Both the values are acceptable, since $	an 	heta = - \frac{4}{3}\,\,$

$\,i.e.,	heta $ lies in ${2^{nd}}$ or ${4^{th}}$ quadrant.

જો $\tan \theta = \frac{{ - 4}}{3},$ તો $\sin \theta = $

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