यदि 90^circ

Question

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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$\sin \,A = \frac{4}{5}$

$\Rightarrow$ $	an A =  - \frac{4}{3}$, $({90^o} < A < {180^o})$

$	an A = \frac{{2	an \frac{A}{2}}}{{1 - {{	an }^2}\frac{A}{2}}}$, (माना $	an \frac{A}{2} = P$)

$ - \frac{4}{3} = \frac{{2P}}{{1 - {P^2}}}$

$\Rightarrow$  $4{P^2} - 6P - 4 = 0$

$\Rightarrow$ $P = \frac{{ - 1}}{2}\,$(असम्भव),

अत:  $	an \frac{A}{2} = 2$.

यदि $90^\circ < A < 180^\circ $ तथा $\sin A = \frac{4}{5},$ तब $\tan \frac{A}{2}$ का मान होगा

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