#### Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 7 sub question (i) maths textbook solution

Answer:  $\sec ^{-1} \frac{x}{a}$

Hint: The range of principal value of   $\cot ^{-1} \text { is }[0, \pi]$
Given:  $\cot ^{-1}\left\{\frac{a}{\sqrt{x^{2}-a^{2}}}\right\},|x|>a$

Explanation:

Let $x=a \sec \theta$

Now, we put value of  $x$  in given question.

\begin{aligned} &\cot ^{-1}\left\{\frac{a}{\sqrt{(a \sec \theta)^{2}-a^{2}}}\right\} \\ &\cot ^{-1}\left\{\frac{a}{\sqrt{a^{2} \sec ^{2} \theta-a^{2}}}\right\} \\ &\cot ^{-1}\left\{\frac{a}{\sqrt{a^{2}\left(\sec ^{2} \theta-1\right)}}\right\} \end{aligned}

As we know   $1+\tan ^{2} \theta=\sec ^{2} \theta$

\begin{aligned} &\sec ^{2} \theta-1=\tan ^{2} \theta \\ &\cot ^{-1}\left\{\frac{a}{\sqrt{a^{2} \tan ^{2} \theta}}\right\} \end{aligned}

Now we remove square root

\begin{aligned} &\cot ^{-1}\left\{\frac{a}{a \tan \theta}\right\} \\ &\cot ^{-1}\left\{\frac{1}{\tan \theta}\right\} \end{aligned}

As we know,   $\cot \theta=\frac{1}{\tan \theta}$

$\cot ^{-1}(\cot \theta)=\theta$

As we know,   $\cot ^{-1}(\cot \theta), \theta \in[0, \pi]$

\begin{aligned} &x=a \sec \theta \\ &\theta=\sec ^{-1} \frac{x}{a} \end{aligned}

Hence, $\cot ^{-1}\left\{\frac{a}{\sqrt{x^{2}-a^{2}}}\right\},|x|>a \text { is } \sec ^{-1} \frac{x}{a}$