How do you simplify the expression $\frac{{tan}^{2} t + 1}{1 + {cot}^{2} t}$ $(tan^2t+1)/(1+cot^2t)$ ?

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Answer 1 · 2016-08-29T06:54:19

${tan}^{2} t$ $tan^2 t$ .

The Exp. $= \frac{{tan}^{2} t + 1}{1 + {cot}^{2} t}$ $=(tan^2 t+1)/(1+cot^2 t)$

$= \frac{\frac{{sin}^{2} t}{{cos}^{2} t} + 1}{1 + \frac{{cos}^{2} t}{{sin}^{2} t}}$ $=(sin^2 t/cos^2 t+1)/(1+cos^2 t/sin^2 t)$

$={(cancel(sin^2 t+cos^2 t))/cos^2 t}/{cancel(cos^2 t+sin^2 t)/sin^2 t}$

$=sin^2 t/cos^2 t$

$=tan^2 t$ .

Alternatively, to shorten the procedure of simplification, we can, use

the Identities $: tan^2 t+1=sec^2 t, and, 1+cot^2 t=csc^2 t$ .

How do you simplify the expression (tan^2t+1)/(1+cot^2t)tan2t+11+cot2t(tan^2t+1)/(1+cot^2t)?