If tan theta = t, then tan 2theta + sec 2thet | vedclass

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Question

(a) $	an 2	heta = \frac{{2	an 	heta }}{{1 - {{	an }^2}	heta }},\cos 2	heta = \frac{{1 - {{	an }^2}	heta }}{{1 + {{	an }^2}	heta }}$

$	a

Vedclass · Accepted Answer

$\frac{{1 + t}}{{1 - t}}$

Vedclass · Answer

(a) $	an 2	heta = \frac{{2	an 	heta }}{{1 - {{	an }^2}	heta }},\cos 2	heta = \frac{{1 - {{	an }^2}	heta }}{{1 + {{	an }^2}	heta }}$

$	an 2	heta + \sec 2	heta = \frac{{2t}}{{1 - {t^2}}} + \frac{{1 + {t^2}}}{{1 - {t^2}}} $

$= \frac{{{{(1 + t)}^2}}}{{(1 - t)(1 + t)}} = \frac{{1 + t}}{{1 - t}}$.

If $\tan \theta = t,$ then $\tan 2\theta + \sec 2\theta = $

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