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3.Trigonometrical Ratios, Functions and Identities
easy
If $90^\circ < A < 180^\circ $ and $\sin A = \frac{4}{5},$ then $\tan \frac{A}{2}$ is equal to
A
$1/2$
B
$3/5$
C
$3/2$
D
$2$
Solution
(d) $\sin \,A = \frac{4}{5}$
==>$\tan A = – \frac{4}{3}$, $({90^o} < A < {180^o})$
$\tan A = \frac{{2\tan \frac{A}{2}}}{{1 – {{\tan }^2}\frac{A}{2}}}$, (Let $\tan \frac{A}{2} = P$)
==> $ – \frac{4}{3} = \frac{{2P}}{{1 – {P^2}}}$
==> $4{P^2} – 6P – 4 = 0$
==> $P = \frac{{ – 1}}{2}{\rm{ (impossible),}}\,$
hence $\tan \frac{A}{2} = 2$.
Standard 11
Mathematics
