Question

If we consider only the principle values of the inverse trigonometric functions, then the values of $\tan \left( {{{\cos }^{ - 1}}\dfrac{1}{{5\sqrt 2 }} - {{\sin }^{ - 1}}\dfrac{4}{{\sqrt {17} }}} \right)$ is
A) $\sqrt {\dfrac{{29}}{3}}$
B) $\dfrac{{29}}{3}$
C) $\sqrt {\dfrac{3}{{29}}}$
D) $\dfrac{3}{{29}}$

Answer 1

Firstly, convert ${\cos ^{ - 1}}\dfrac{1}{{5\sqrt 2 }}$ and ${\sin ^{ - 1}}\dfrac{4}{{\sqrt {17} }}$ in the terms of ${\tan ^{ - 1}}$.
After that, substitute the values in terms of ${\tan ^{ - 1}}$ , apply the property ${\tan ^{ - 1}}a - {\tan ^{ - 1}}b = {\tan ^{ - 1}}\left( {\dfrac{{a - b}}{{1 + ab}}} \right)$.
Thus, get the answer and choose the correct option.

Complete step by step solution:
We are given that $\tan \left( {{{\cos }^{ - 1}}\dfrac{1}{{5\sqrt 2 }} - {{\sin }^{ - 1}}\dfrac{4}{{\sqrt {17} }}} \right)$ .
To solve it, first we have to convert ${\cos ^{ - 1}}\dfrac{1}{{5\sqrt 2 }}$ and ${\sin ^{ - 1}}\dfrac{4}{{\sqrt {17} }}$ in the terms of ${\tan ^{ - 1}}$ .

$${\cos ^{ - 1}}\dfrac{1}{{5\sqrt 2 }} = {\tan ^{ - 1}}\dfrac{{\sqrt {{{\left( {5\sqrt 2 } \right)}^2} - {{\left( 1 \right)}^2}} }}{1} \\\ = {\tan ^{ - 1}}\dfrac{{\sqrt {50 - 1} }}{1} \\\ = {\tan ^{ - 1}}\sqrt {49} \\\ = {\tan ^{ - 1}}7$$

And
${\sin ^{ - 1}}\dfrac{4}{{\sqrt {17} }} = {\tan ^{ - 1}}\dfrac{4}{{\sqrt {{{\left( {\sqrt {17} } \right)}^2} - {{\left( 4 \right)}^2}} }} \\\ = {\tan ^{ - 1}}\dfrac{4}{{\sqrt {17 - 16} }} \\\ = {\tan ^{ - 1}}\dfrac{4}{{\sqrt 1 }} \\\ = {\tan ^{ - 1}}4$
Now, substituting ${\cos ^{ - 1}}\dfrac{1}{{5\sqrt 2 }} = {\tan ^{ - 1}}7$ and ${\sin ^{ - 1}}\dfrac{4}{{\sqrt {17} }} = {\tan ^{ - 1}}4$ in $\tan \left( {{{\cos }^{ - 1}}\dfrac{1}{{5\sqrt 2 }} - {{\sin }^{ - 1}}\dfrac{4}{{\sqrt {17} }}} \right)$ .
$\therefore \tan \left( {{{\cos }^{ - 1}}\dfrac{1}{{5\sqrt 2 }} - {{\sin }^{ - 1}}\dfrac{4}{{\sqrt {17} }}} \right) = \tan \left( {{{\tan }^{ - 1}}7 - {{\tan }^{ - 1}}4} \right)$
Then, applying the property ${\tan ^{ - 1}}a - {\tan ^{ - 1}}b = {\tan ^{ - 1}}\left( {\dfrac{{a - b}}{{1 + ab}}} \right)$ .
$\therefore \tan \left( {{{\tan }^{ - 1}}7 - {{\tan }^{ - 1}}4} \right) = \tan \left( {{{\tan }^{ - 1}}\left( {\dfrac{{7 - 4}}{{1 + 7 \times 4}}} \right)} \right)$
$= \tan \left( {{{\tan }^{ - 1}}\left( {\dfrac{3}{{1 + 28}}} \right)} \right) \\\ = \tan \left( {{{\tan }^{ - 1}}\left( {\dfrac{3}{{29}}} \right)} \right) \\\ = \dfrac{3}{{29}} \\\$
Thus, $\tan \left( {{{\cos }^{ - 1}}\dfrac{1}{{5\sqrt 2 }} - {{\sin }^{ - 1}}\dfrac{4}{{\sqrt {17} }}} \right) = \dfrac{3}{{29}}$ .

So, option (D) is correct.

Note:
Some properties of inverse trigonometric tan functions:

\tan \left( {{{\tan }^{ - 1}}x} \right) = x, - \infty \leqslant x \leqslant \infty \\\ {\tan ^{ - 1}}\left( {\tan x} \right) = x, - \dfrac{\pi }{2} \leqslant x \leqslant \dfrac{\pi }{2} \\\ {\tan ^{ - 1}}\left( { - x} \right) = - {\tan ^{ - 1}}x \\\ {\tan ^{ - 1}}x + {\cot ^{ - 1}}x = 1 \\\ {\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \left\\{ {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right),xy < 1 \\\ \pi + {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right),xy > 1 \\\ \right. \\\ {\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right) \\\