#### Explain solution RD Sharma class 12 chapter Inverse Trigonometric Function exercise 3.9 question 2 maths subquestion (i)

$\frac{-24}{7}$

Hint:

Get rid of the negative sign by using the formula

$cos^{-1}(-x)=\pi -cos^{-1}(x)$

Convert the cos-1 term to a sec-1 term.

Now use the formula

$sec^{2}(x)=1+tan^{2}(x)$

$tan(x)=\sqrt{sec^{2}(x)-1}$

Concept:

Inverse Trigonometry

Solution:

$tan(cos^{-1}(\frac{-7}{25}))=tan(\pi -cos^{-1}(\frac{7}{25}))$                    $[cos^{-1}(-x)=\pi -cos^{-1}(x)]$

$=-tan(\pi -cos^{-1}(\frac{7}{25}))$                $[tan(\pi -\theta )=-tan\theta ]$

$=-tan(sec^{-1}(\frac{25}{7}))$                         $[cos^{-1}(x)=sec^{-1}(\frac{1}{x})]$

$=-\sqrt{sec^{2}(sec^{-1}(\frac{25}{7}))-1}$           $[tan^{2}\theta =sec^{2}\theta -1]$

$=-\sqrt{\frac{625-49}{49}}$              $=-\frac{24}{7}$

Note:   Try to convert the inner inverse term to the inverse of outer trigo ratio or the trigo ratio that can be related directly to the outer ratio using identities