#### Need solution for RD Sharma maths class 12 chapter Inverse Trigonometric Function exercise 3.9 question 1 subquestion (iii)

$\frac{-5}{12}$

Hint:

Eliminate the negative sign by using formula sec-1(-x)=$\pi$-sec-1(x) and then convert the sec-1 term into a cot-1 term to simplify the sum

Concept:

Inverse Trigonometric

Solution:

$cot(sec^{-1}(\frac{-13}{5}))=cot(\pi -sec^{-1}(\frac{13}{5}))$                        $[sec^{-1}(-x)=\pi -sec^{-1}(x)]$

$=-cot(sec^{-1}(\frac{13}{5}))$                              $[cot(\pi -\theta) =-cot\, \theta ]$

Let

$sec^{-1}(\frac{13}{5})=x$                    ..............(1)

$sec\, x=\frac{13}{5}$

$\therefore tan\, x=\sqrt{sec^{2}x-1}$

$=\sqrt{(\frac{13}{5})^{2}-1}\: \: \: =\frac{12}{5}$

$cot\, x=\frac{5}{12}\Rightarrow x=cot^{-1}(\frac{5}{12})$                                        ............(2)

$sec^{-1}(\frac{13}{5})=cot^{-1}(\frac{5}{12})$

From (1) and (2)

$=cot(sec^{-1}(\frac{-13}{5}))$

$=-cot(sec^{-1}(\frac{13}{5}))$

$=-cot(cot(\frac{5}{12}))$

$=\frac{-5}{12}$

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

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