If 90^circ

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Question

(d) $\sin \,A = \frac{4}{5}$

==>$	an A = - \frac{4}{3}$, $({90^o} < A < {180^o})$

$	an A = \frac{{2	an \frac{A}{2}}}{{1 - {{	an }^2

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(d) $\sin \,A = \frac{4}{5}$ ==>$ an A = - \frac{4}{3}$, $({90^o} < A < {180^o})$ $ an A = \frac{{2 an \frac{A}{2}}}{{1 - {{ an }^2}\frac{A}{2}}}$, (Let $ an \frac{A}{2} = P$) ==> $ - \frac{4}{3} = \frac{{2P}}{{1 - {P^2}}}$ ==> $4{P^2} - 6P - 4 = 0$ ==> $P = \frac{{ - 1}}{2}{ m{ (impossible),}}\,$ hence $ an \frac{A}{2} = 2$.

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

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