^ - 1 [ ( )^ 1/2 ] - ^ - 1 [ ( )^ 1/2 ] = x, then x = - Vidyadip

MCQ

${\cot ^{ - 1}}[{(\cos \alpha )^{1/2}}] - {\tan ^{ - 1}}[{(\cos \alpha )^{1/2}}] = x,$ then $\sin x = $

✓
${\tan ^2}\left( {\frac{\alpha }{2}} \right)$
B
${\cot ^2}\left( {\frac{\alpha }{2}} \right)$
C
$\tan \alpha $
D
$\cot \left( {\frac{\alpha }{2}} \right)$

✓

Answer

Correct option: A.

${\tan ^2}\left( {\frac{\alpha }{2}} \right)$

a
(a) ${\tan ^{ - 1}}\left[ {\frac{1}{{\sqrt {\cos \alpha } }}} \right] - {\tan ^{ - 1}}\left[ {\sqrt {\cos \alpha } } \right] = x$

==>${\tan ^{ - 1}}\left[ {\frac{{\frac{1}{{\sqrt {\cos \alpha } }} - \sqrt {\cos \alpha } }}{{1 + \frac{{\sqrt {\cos \alpha } }}{{\sqrt {\cos \alpha } }}}}} \right] = x$

==> $\tan x = \frac{{1 - \cos \alpha }}{{2\sqrt {\cos \alpha } }}$

$\therefore \sin x = \frac{{1 - \cos \alpha }}{{1 + \cos \alpha }} = \frac{{2{{\sin }^2}\frac{\alpha }{2}}}{{2{{\cos }^2}\frac{\alpha }{2}}} = {\tan ^2}\left( {\frac{\alpha }{2}} \right)$.

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