Question

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

Answer 1

Hint: Convert $${\cos }^{-1}{\displaystyle \frac{1}{\sqrt{2}}}$$ into $${\tan }^{-1}$$ function by assuming 1 as base and $$\sqrt{2}$$ as hypotenuse. Also convert $${\sin }^{-1}{\displaystyle \frac{4}{\sqrt{17}}}$$ 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: - $${\tan }^{-1}a-{\tan }^{-1}b={\tan }^{-1}\left({\displaystyle \frac{a-b}{1+ab}}\right)$$ 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: -
$$\Rightarrow E=\tan [{\cos }^{-1}{\displaystyle \frac{1}{\sqrt{2}}}-{\sin }^{-1}{\displaystyle \frac{4}{\sqrt{17}}}]$$
Let us consider $${\cos }^{-1}$$ and $${\sin }^{-1}$$ functions into $${\tan }^{-1}$$ function.
We know that, $$\cos \theta$$ = ${\displaystyle \frac{\text{Base}}{\text{Hypotenuse}}}$ $$\Rightarrow \theta ={\cos }^{-1}$$$({\displaystyle \frac{\text{Base}}{\text{Hypotenuse}}})$
On comparing the above relation with $${\cos }^{-1}\left({\displaystyle \frac{1}{\sqrt{2}}}\right)$$, 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.
$$\begin{array}{rl}& \Rightarrow {\left(\sqrt{2}\right)}^2=p^2+1^2\\ & \Rightarrow 2=p^2+1\\ & \Rightarrow p^2=1\\ & \Rightarrow p=1\end{array}$$
Hence, $${\cos }^{-1}\left({\displaystyle \frac{1}{\sqrt{2}}}\right)={\tan }^{-1}\left({\displaystyle \frac{p}{b}}\right)={\tan }^{-1}\left({\displaystyle \frac{1}{1}}\right)={\tan }^{-1}1$$.
Now, we know that, $$\sin \theta ={\displaystyle \frac{p}{h}}\Rightarrow \theta ={\sin }^{-1}\left({\displaystyle \frac{p}{h}}\right)$$.
On comparing the above relation with $${\sin }^{-1}\left({\displaystyle \frac{4}{\sqrt{17}}}\right)$$, we have, p = 4, h = $$\sqrt{17}$$.
Therefore, applying Pythagoras theorem, we get,
$$\begin{array}{rl}& \Rightarrow h^2=p^2+b^2\\ & \Rightarrow {\left(\sqrt{17}\right)}^2=4^2+b^2\\ & \Rightarrow 17=16+b^2\\ & \Rightarrow b^2=1\\ & \Rightarrow b=1\end{array}$$
Hence, $${\sin }^{-1}\left({\displaystyle \frac{4}{\sqrt{17}}}\right)={\tan }^{-1}\left({\displaystyle \frac{p}{b}}\right)={\tan }^{-1}\left({\displaystyle \frac{4}{1}}\right)={\tan }^{-1}4$$.
So, the expression becomes: -
$$\Rightarrow E=\tan [{\tan }^{-1}1-{\tan }^{-1}4]$$
Applying the identity: - $${\tan }^{-1}a-{\tan }^{-1}b={\tan }^{-1}\left({\displaystyle \frac{a-b}{1+ab}}\right)$$, we get,
$$\begin{array}{rl}& \Rightarrow E=\tan \left[{\tan }^{-1}\left({\displaystyle \frac{1-4}{1+1\times 4}}\right)\right]\\ & \Rightarrow E=\tan \left[{\tan }^{-1}\left({\displaystyle \frac{-3}{5}}\right)\right]\end{array}$$
Finally using the identity, $$\tan \left[{\tan }^{-1}x\right]=x$$, we have,
$$\Rightarrow E={\displaystyle \frac{-3}{5}}$$
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 $${\cos }^{-1}\left({\displaystyle \frac{1}{\sqrt{2}}}\right)={\tan }^{-1}1$$. Here, we knew that $${\cos }^{-1}\left({\displaystyle \frac{\pi }{4}}\right)={\displaystyle \frac{1}{\sqrt{2}}}$$. So, $${\displaystyle \frac{\pi }{4}}={\cos }^{-1}\left({\displaystyle \frac{1}{\sqrt{2}}}\right)$$. Hence, the angle was $${\displaystyle \frac{\pi }{4}}$$ and $$\tan \left({\displaystyle \frac{\pi }{4}}\right)=1$$. So, we did not required Pythagoras theorem here. But for the conversion of $${\sin }^{-1}\left({\displaystyle \frac{4}{\sqrt{17}}}\right)$$ into $${\tan }^{-1}$$ function we needed Pythagoras theorem because we don’t know any particular angle for that.