If tan A = frac{{1 - cos B}}{{sin B}}, find tan 2A in terms of tan B and show that

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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

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