How do you find the exact value of $cos(tan^-1 (5/12)+cot^-1 (5/12))$ ?

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Answer 1 · 2016-02-25T16:08:01

$cos(tan^(−1)(5/12)+cot^(−1)(5/12))=cos(pi/2)=0$

Remember the trigonometric identity

$tantheta=cot(pi/2-theta)=x$ i.e.

$tan^-1x=theta$ and $cot^-1x=(pi/2-theta)$

Hence, for any $x$ , $tan^(−1)x+cot^(−1)x=(theta+pi/2-theta)=pi/2$

Hence for any $x$ , $cos(tan^(−1)x+cot^(−1)x)=cos(pi/2)=0$

Note that $x$ is equal to $tantheta$ , $x$ can take any value ${-oo,oo}$ (as this is the range of $tantheta$ ) including $5/12$ .

How do you find the exact value of cos(tan^-1 (5/12)+cot^-1 (5/12))cos(tan^-1 (5/12)+cot^-1 (5/12))?