frac{{2sin theta ,tan theta (1 - tan theta ) | vedclass

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Question

(b) Given expression

$ = \frac{{2\,\sin 	heta }}{{{{(1 + 	an \,	heta )}^2}}}\,\left\{ {	an \,	heta \,(1 - 	an \,	heta ) + {{\sec }^2}	heta

Vedclass · Accepted Answer

$\frac{{2\,\sin 	heta }}{{1 + 	an 	heta }}$

Vedclass · Answer

(b) Given expression

$ = \frac{{2\,\sin 	heta }}{{{{(1 + 	an \,	heta )}^2}}}\,\left\{ {	an \,	heta \,(1 - 	an \,	heta ) + {{\sec }^2}	heta } ight\}$

$ = \frac{{2\,\sin 	heta }}{{{{(1 + 	an \,	heta )}^2}}}\,\left\{ {	an \,	heta \, - {{	an }^2}\,	heta + 1 + {{	an }^2}	heta } ight\}$

$ = \frac{{2\,\sin 	heta }}{{1 + 	an 	heta }}$.

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

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