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# 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) $\dfrac{-3}{5}$

Last updated date: 16th Sep 2024
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Hint: 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.
\begin{align} & \Rightarrow {{\left( \sqrt{2} \right)}^{2}}={{p}^{2}}+{{1}^{2}} \\ & \Rightarrow 2={{p}^{2}}+1 \\ & \Rightarrow {{p}^{2}}=1 \\ & \Rightarrow p=1 \\ \end{align}
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{align} & \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{align}
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{align} & \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{align}
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.