Question

If we consider only the principal values of the inverse trigonometric functions, then the value of $\tan \left({\cos }^{-1}\frac{1}{5\sqrt{2}}-{\sin }^{-1}\frac{4}{\sqrt{17}}\right)$.

• A. $\sqrt{\frac{29}{3}}$
• B. $\frac{29}{3}$
• C. $\sqrt{\frac{3}{29}}$
• D. $\frac{3}{29}$

Answer 1

The correct option is D

$\frac{3}{29}$

Explanation for the correct option:

Finding the value of $\mathit{t}\mathit{a}\mathit{n}\mathbf{(}\mathbf{c}\mathbf{o}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}\frac{\mathbf{1}}{\mathbf{5}\sqrt{\mathbf{2}}}\mathbf{-}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{-}\mathbf{1}}\frac{\mathbf{4}}{\sqrt{\mathbf{17}}}\mathbf{)}$:

Let, ${\cos }^{-1}\frac{1}{5\sqrt{2}}=A$.

$$\begin{gathered} \cos A=\frac{1}{5\sqrt{2}} \\ \therefore \tan A=7 \end{gathered}$$

Thus ${\cos }^{-1}\frac{1}{5\sqrt{2}}$ can be written as ${\tan }^{-1}7$.

Similarly,

Let, ${\sin }^{-1}\frac{4}{\sqrt{17}}=B$.

$$\begin{gathered} \sin B=\frac{4}{\sqrt{17}} \\ \therefore \tan B=4 \end{gathered}$$

Thus ${\sin }^{-1}\frac{4}{\sqrt{17}}$ can be written as ${\tan }^{-1}4$.

Now,

$\begin{array}{rcl}tan({\cos }^{-1}\frac{1}{5\sqrt{2}}-{\sin }^{-1}\frac{4}{\sqrt{17}})& =& \tan \left({\tan }^{-1}7-{\tan }^{-1}4\right)\\ & =& \frac{\tan \left({\tan }^{-1}7\right)-\tan \left({\tan }^{-1}4\right)}{1+\tan \left({\tan }^{-1}7\right)\tan \left({\tan }^{-1}4\right)}\left[\mathbf{\because }\mathit{t}\mathit{a}\mathit{n}\mathbf{(}\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{)}\mathbf{=}\frac{\mathbf{tan}\mathbf{}\mathbf{a}\mathbf{-}\mathbf{tan}\mathbf{}\mathbf{b}}{\mathbf{1}\mathbf{+}\mathbf{tan}\mathbf{}\mathbf{a}\mathbf{}\mathbf{tan}\mathbf{}\mathbf{b}}\right]\\ & =& \frac{7-4}{1+28}\\ & =& \frac{3}{29}\end{array}$

Hence, the correct option is D.