How do you find the exact value of $sec(tan^-1(-3/5))$ ?

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Answer 1 · 2017-01-04T08:17:07

$sec(tan^(-1)(-3/5))=sqrt34/5$

Let $theta=tan^(-1)(-3/5)$

then $tantheta=-3/5$ and

$sec^2theta=1+tan^2theta=1+9/25=34/25$

As $tantheta$ is negative, as range of $tan^(-1)$ is $-pi/2< theta < pi/2$ , $theta$ lies in Q4 and $sectheta$ is positive and $sectheta=sqrt34/5$

and hence $sec(tan^(-1)(-3/5))$

= $sectheta$

= $sqrt34/5$

How do you find the exact value of sec(tan^-1(-3/5))sec(tan^-1(-3/5))?