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

$\frac{13}{5}$

Hint:

Convert the cot-1 term to a tan-1 term so that we can us the following result to solve the sum

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

$\therefore sec\, (x)=\sqrt{1+tan^{2}(x)}$            ...using this we can simplify the sum

Concept:

Inverse Trigonometry

Solution:

$sec[cot^{-1}(\frac{-5}{12})]=sec[tan^{-1}(-\frac{12}{5})]$                        $[cot^{-1}(x)=tan^{-1}(\frac{1}{x})]$

$=sec[-tan^{-1}(\frac{12}{5})]$                      $[tan^{-1}(-x)=-tan^{-1}(x)]$

$=sec[tan^{-1}(\frac{12}{5})]$                         $[sec(-\theta )=sec\, \theta ]$

$=\sqrt{1+tan^{2}(tan^{-1}(\frac{12}{5}))}$        $[sec^{2}\theta =1+tan^{2}\theta ]$

$=\sqrt{1+(\frac{12}{5})^{2}}\; \; \; =>\sqrt{1+\frac{144}{25}}\; \; \; =\sqrt{\frac{25+144}{25}}$

$=\sqrt{\frac{169}{25}}$

$=\frac{13}{5}$

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