Question

If we consider only the principal values of the inverse trigonometric functions, then the value of $\tan \left[ {{\cos }^{-1}}\dfrac{1}{\sqrt{2}}-{{\sin }^{-1}}\dfrac{4}{\sqrt{17}} \right]$ is: -
(a) $\dfrac{\sqrt{29}}{3}$
(b) $\dfrac{29}{3}$
(c) $\dfrac{\sqrt{3}}{29}$
(d) $\dfrac{-3}{5}$

Answer 1

Convert ${{\cos }^{-1}}\dfrac{1}{\sqrt{2}}$ into ${{\tan }^{-1}}$ function by assuming 1 as base and $\sqrt{2}$ as hypotenuse. Also convert ${{\sin }^{-1}}\dfrac{4}{\sqrt{17}}$ into ${{\tan }^{-1}}$ function by assuming 4 as perpendicular and $\sqrt{17}$ as hypotenuse. Use Pythagoras theorem given by: - ${{h}^{2}}={{p}^{2}}+{{b}^{2}}$ for the above two process. Here, h = hypotenuse, p = perpendicular and b = base. Now, apply the identity: - ${{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)$ to simplify. Finally apply the rule: - $\tan \left( {{\tan }^{-1}}x \right)=x$ to get the answer.

Complete step-by-step solution
We have been provided with the expression: -
$\Rightarrow E=\tan \left[ {{\cos }^{-1}}\dfrac{1}{\sqrt{2}}-{{\sin }^{-1}}\dfrac{4}{\sqrt{17}} \right]$
Let us consider ${{\cos }^{-1}}$ and ${{\sin }^{-1}}$ functions into ${{\tan }^{-1}}$ function.
We know that, $\cos \theta$ = $\dfrac{\text{Base}}{\text{Hypotenuse}}$ \Rightarrow \theta ={{\cos }^{-1}}$$$(\dfrac{\text{Base}}{\text{Hypotenuse}})$ On comparing the above relation with {{\cos }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)$, we get, base = 1, hypotenuse =$\sqrt{2}$. Therefore, applying Pythagoras theorem, we get,${{h}^{2}}={{p}^{2}}+{{b}^{2}}$$, where h = hypotenuse, p = perpendicular and b = base.

& \Rightarrow {{\left( \sqrt{2} \right)}^{2}}={{p}^{2}}+{{1}^{2}} \\\ & \Rightarrow 2={{p}^{2}}+1 \\\ & \Rightarrow {{p}^{2}}=1 \\\ & \Rightarrow p=1 \\\ \end{aligned}$$ Hence, $${{\cos }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)={{\tan }^{-1}}\left( \dfrac{p}{b} \right)={{\tan }^{-1}}\left( \dfrac{1}{1} \right)={{\tan }^{-1}}1$$. Now, we know that, $$\sin \theta =\dfrac{p}{h}\Rightarrow \theta ={{\sin }^{-1}}\left( \dfrac{p}{h} \right)$$. On comparing the above relation with $${{\sin }^{-1}}\left( \dfrac{4}{\sqrt{17}} \right)$$, we have, p = 4, h = $$\sqrt{17}$$. Therefore, applying Pythagoras theorem, we get, $$\begin{aligned} & \Rightarrow {{h}^{2}}={{p}^{2}}+{{b}^{2}} \\\ & \Rightarrow {{\left( \sqrt{17} \right)}^{2}}={{4}^{2}}+{{b}^{2}} \\\ & \Rightarrow 17=16+{{b}^{2}} \\\ & \Rightarrow {{b}^{2}}=1 \\\ & \Rightarrow b=1 \\\ \end{aligned}$$ Hence, $${{\sin }^{-1}}\left( \dfrac{4}{\sqrt{17}} \right)={{\tan }^{-1}}\left( \dfrac{p}{b} \right)={{\tan }^{-1}}\left( \dfrac{4}{1} \right)={{\tan }^{-1}}4$$. So, the expression becomes: - $$\Rightarrow E=\tan \left[ {{\tan }^{-1}}1-{{\tan }^{-1}}4 \right]$$ Applying the identity: - $${{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)$$, we get, $$\begin{aligned} & \Rightarrow E=\tan \left[ {{\tan }^{-1}}\left( \dfrac{1-4}{1+1\times 4} \right) \right] \\\ & \Rightarrow E=\tan \left[ {{\tan }^{-1}}\left( \dfrac{-3}{5} \right) \right] \\\ \end{aligned}$$ Finally using the identity, $$\tan \left[ {{\tan }^{-1}}x \right]=x$$, we have, $$\Rightarrow E=\dfrac{-3}{5}$$ **Hence, option (d) is the correct answer.** **Note:** One may note that we can decrease the steps of solution a little by directly saying that $${{\cos }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)={{\tan }^{-1}}1$$. Here, we knew that $${{\cos }^{-1}}\left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$$. So, $$\dfrac{\pi }{4}={{\cos }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)$$. Hence, the angle was $$\dfrac{\pi }{4}$$ and $$\tan \left( \dfrac{\pi }{4} \right)=1$$. So, we did not required Pythagoras theorem here. But for the conversion of $${{\sin }^{-1}}\left( \dfrac{4}{\sqrt{17}} \right)$$ into $${{\tan }^{-1}}$$ function we needed Pythagoras theorem because we don’t know any particular angle for that.