1.

If tan \(A\; = \;\frac{{1 - cosB}}{{sinB}},\) then what is \(\frac{{2tanA}}{{1 + {{\tan }^2}A}}\) equal to?1). \(\frac{{sinB}}{2}\)2). 2tanB3). sinB4). 4sinB

Answer»

$(tanA\; = \;\frac{{1 - cosB}}{{sinB}}\; = \;\frac{{2{{\SIN }^2}\frac{B}{2}}}{{2\sin \frac{B}{2}\COS \frac{B}{2}}}\; = \;\tan \frac{B}{2})$

(? cos2x = 1 – 2sin2x & sin2x = 2sinxcosx)

⇒ A = B/2

We know that,

sin 2x = 2tanx/(1 + tan2x)

$(\begin{array}{l}\frac{{2tanA}}{{1 + {{\tan }^2}A}}\; = \;\sin \left( {2\; \TIMES \;\frac{B}{2}} \right)\; = \;\sin B\\\therefore {\rm{\;}}\frac{{2tanA}}{{1 + {{\tan }^2}A}}\; = \;\sin B\end{array})$



Discussion

No Comment Found