Question

Solve the following inverse trigonometric equation equation:
$\cos \left( {{\tan }^{-1}}x \right)=\sin \left( {{\cot }^{-1}}\dfrac{3}{4} \right)$

Answer 1

Hint: For solving this question first we will assume ${{\tan }^{-1}}x=\alpha$ and ${{\cot }^{-1}}\dfrac{3}{4}=\beta$ . After that, we will try to transform the given equation $\cos \alpha =\sin \beta$ in terms of $\tan \alpha$ and $\cot \beta$ with the help of trigonometric formulas like ${{\tan }^{2}}\theta +1={{\sec }^{2}}\theta$ and ${{\cot }^{2}}\theta +1={{\csc }^{2}}\theta$ . Then, we will put $\tan \alpha =x$ and $\cot \beta =\dfrac{3}{4}$ to solve further for the suitable values of $x$ .

Complete step-by-step solution -
Given:
We have to find the suitable values of $x$ from the following equation:
$\cos \left( {{\tan }^{-1}}x \right)=\sin \left( {{\cot }^{-1}}\dfrac{3}{4} \right)$
Now, let ${{\tan }^{-1}}x=\alpha$ and ${{\cot }^{-1}}\dfrac{3}{4}=\beta$ . Then,
$\begin{aligned} & \cos \left( {{\tan }^{-1}}x \right)=\sin \left( {{\cot }^{-1}}\dfrac{3}{4} \right) \\\ & \Rightarrow \cos \alpha =\sin \beta \\\ \end{aligned}$
Now, we will square both terms in the above equation. Then,
$\begin{aligned} & \cos \alpha =\sin \beta \\\ & \Rightarrow {{\cos }^{2}}\alpha ={{\sin }^{2}}\beta \\\ \end{aligned}$
Now, as we know that, $\cos \alpha =\dfrac{1}{\sec \alpha }$ and $\sin \beta =\dfrac{1}{\csc \beta }$ . Then,
$\begin{aligned} & {{\cos }^{2}}\alpha ={{\sin }^{2}}\beta \\\ & \Rightarrow \dfrac{1}{{{\sec }^{2}}\alpha }=\dfrac{1}{{{\csc }^{2}}\beta } \\\ & \Rightarrow {{\csc }^{2}}\beta ={{\sec }^{2}}\alpha .....................\left( 1 \right) \\\ \end{aligned}$
Now, before we proceed we should know the following formulas:
$\begin{aligned} & {{\tan }^{2}}\theta +1={{\sec }^{2}}\theta ..................\left( 2 \right) \\\ & {{\cot }^{2}}\theta +1={{\csc }^{2}}\theta ..................\left( 3 \right) \\\ \end{aligned}$
Now, as per our assumption ${{\tan }^{-1}}x=\alpha$ and ${{\cot }^{-1}}\dfrac{3}{4}=\beta$ . And we know that, $\tan \left( {{\tan }^{-1}}x \right)=x$ and $\cot \left( {{\cot }^{-1}}x \right)=x$ for any $x\in R$ . Then,
$\begin{aligned} & {{\tan }^{-1}}x=\alpha \\\ & \Rightarrow \tan \left( {{\tan }^{-1}}x \right)=\tan \alpha \\\ & \Rightarrow x=\tan \alpha \\\ & \Rightarrow \tan \alpha =x............................\left( 4 \right) \\\ & {{\cot }^{-1}}\dfrac{3}{4}=\beta \\\ & \Rightarrow \cot \left( {{\cot }^{-1}}\dfrac{3}{4} \right)=\cot \beta \\\ & \Rightarrow \dfrac{3}{4}=\cot \beta \\\ & \Rightarrow \cot \beta =\dfrac{3}{4}...........................\left( 5 \right) \\\ \end{aligned}$
Now, we will use the formula from the equation (2) to write ${{\sec }^{2}}\alpha =1+{{\tan }^{2}}\alpha$ in equation (1) and formula from equation (3) to write ${{\csc }^{2}}\beta ={{\cot }^{2}}\beta +1$ in equation (1). Then,
$\begin{aligned} & {{\csc }^{2}}\beta ={{\sec }^{2}}\alpha \\\ & \Rightarrow 1+{{\cot }^{2}}\beta =1+{{\tan }^{2}}\alpha \\\ & \Rightarrow {{\cot }^{2}}\beta ={{\tan }^{2}}\alpha \\\ \end{aligned}$
Now, we will put $\tan \alpha =x$ from equation (4) and $\cot \beta =\dfrac{3}{4}$ from equation (5) in the above equation. Then,
$\begin{aligned} & {{\cot }^{2}}\beta ={{\tan }^{2}}\alpha \\\ & \Rightarrow {{\left( \dfrac{3}{4} \right)}^{2}}={{x}^{2}} \\\ & \Rightarrow {{x}^{2}}={{\left( \dfrac{3}{4} \right)}^{2}} \\\ & \Rightarrow x=\pm \dfrac{3}{4} \\\ \end{aligned}$
Now, from the above result we conclude that, if $\cos \left( {{\tan }^{-1}}x \right)=\sin \left( {{\cot }^{-1}}\dfrac{3}{4} \right)$ then, the value of $x$ will be equal to $\dfrac{3}{4},-\dfrac{3}{4}$ . Then,
$\begin{aligned} & \cos \left( {{\tan }^{-1}}x \right)=\sin \left( {{\cot }^{-1}}\dfrac{3}{4} \right) \\\ & \Rightarrow x=\pm \dfrac{3}{4} \\\ \end{aligned}$
Thus, suitable values of $x$ will be $x=\pm \dfrac{3}{4}$ .

Note: Here, the student should first understand what is asked in the question, and then proceed in the right direction to get the correct answer quickly. After that, we should proceed in a stepwise manner in such questions and apply basic formulas of trigonometry for better clarity. And avoid calculation error while solving. Moreover, we could have directly calculated the value of $\cos \left( {{\tan }^{-1}}x \right)$ by formula $\cos \left( {{\tan }^{-1}}x \right)=\dfrac{1}{\sqrt{1+{{x}^{2}}}}$ , where $x\in R$ and value of $\sin \left( {{\cot }^{-1}}\dfrac{3}{4} \right)$ by formula $\sin \left( {{\cot }^{-1}}x \right)=\dfrac{1}{\sqrt{1+{{x}^{2}}}}$ , where $x\in R$ .