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

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.