Question

If $\alpha ={\cos }^{-1}\left(\frac{3}{5}\right),\beta ={\tan }^{-1}\left(\frac{1}{3}\right),$ where $0<\alpha ,\beta <\frac{\pi }{2}$, then $\alpha -\beta$ is equal to:

• A. ${\tan }^{-1}\left(\frac{9}{5\sqrt{10}}\right)$
• B. ${\sin }^{-1}\left(\frac{9}{5\sqrt{10}}\right)$
• C. ${\tan }^{-1}\left(\frac{9}{15}\right)$
• D. ${\cos }^{-1}\left(\frac{9}{5\sqrt{10}}\right)$

Answer 1

The correct option is B

${\sin }^{-1}\left(\frac{9}{5\sqrt{10}}\right)$

Find the value of $\alpha -\beta$:

Given,

$\begin{array}{rcl}\alpha & =& {\cos }^{-1}\left(\frac{3}{5}\right)\\ & =& {\tan }^{-1}\left(\frac{4}{3}\right)\end{array}$

$\beta ={\tan }^{-1}\left(\frac{1}{3}\right)$

$\therefore \alpha -\beta ={\tan }^{-1}\left(\frac{4}{3}\right)–{\tan }^{-1}\left(\frac{1}{3}\right)$

$={\tan }^{-1}\left(\frac{{\displaystyle \frac{4}{3}-\frac{1}{3}}}{1+\frac{4}{3}\times \frac{1}{3}}\right)$ $\left[\mathbf{\because }\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{A}\mathbf{}\mathbf{-}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{B}\mathbf{}\mathbf{=}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{A-B}{1+AB}\right)\right]$

$={\tan }^{-1}\left(\frac{1}{1+{\displaystyle \frac{4}{9}}}\right)$

$={\tan }^{-1}\left(\frac{9}{13}\right)$

$={\sin }^{-1}\left(\frac{9}{\sqrt{{13}^2+9^2}}\right)$ $\left[\mathbf{\because }\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{A}{B}\right)\mathbf{}\mathbf{=}\mathbf{}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{1}{\sqrt{A^2+B^2}}\right)\right]$

$={\sin }^{-1}\left(\frac{9}{\sqrt{250}}\right)$

$=\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{9}{5\sqrt{10}}\right)$

Hence, Option ‘B’ is Correct.