Frac{{2sin θ ,tan θ (1 - tan θ ) + 2sin θ {{sec }^2}θ }}{{{{(1 + tan θ )}^2}}} =

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3.Trigonometrical Ratios, Functions and Identities

easy

$\frac{{2\sin \theta \,\tan \theta (1 - \tan \theta ) + 2\sin \theta {{\sec }^2}\theta }}{{{{(1 + \tan \theta )}^2}}} = $

$\frac{{\sin \,\theta }}{{1 + \tan \theta }}$

$\frac{{2\,\sin \theta }}{{1 + \tan \theta }}$

$\frac{{2\sin \theta }}{{{{(1 + \tan \theta )}^2}}}$

None of these

(b) Given expression

$ = \frac{{2\,\sin \theta }}{{{{(1 + \tan \,\theta )}^2}}}\,\left\{ {\tan \,\theta \,(1 – \tan \,\theta ) + {{\sec }^2}\theta } \right\}$

$ = \frac{{2\,\sin \theta }}{{{{(1 + \tan \,\theta )}^2}}}\,\left\{ {\tan \,\theta \, – {{\tan }^2}\,\theta + 1 + {{\tan }^2}\theta } \right\}$

$ = \frac{{2\,\sin \theta }}{{1 + \tan \theta }}$.

Standard 11

Mathematics

normal

(KVPY-2010)

medium

easy

easy

medium