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3.Trigonometrical Ratios, Functions and Identities
easy
If $\tan A = \frac{{1 - \cos B}}{{\sin B}},$ find $\tan 2A$ in terms of $\tan B$ and show that
A
$\tan 2A = \tan B$
B
$\tan 2A = {\tan ^2}B$
C
$\tan 2A = {\tan ^2}B + 2\tan B$
D
None of the above
(IIT-1983)
Solution
(a) $\tan A = \frac{{1 – \cos B}}{{\sin B}}$
$ = \frac{{2{{\sin }^2}(B/2)}}{{2\sin (B/2)\cos (B/2)}} = \tan \frac{B}{2}$
==> $\tan 2A = \tan B$.
Standard 11
Mathematics