Question

$2\pi -\left({\sin }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)+{\sin }^{-1}\left(\frac{16}{65}\right)\right)$is equal to

• A. $\frac{5\pi }{4}$
• B. $\frac{3\pi }{2}$
• C. $\frac{7\pi }{4}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B

$\frac{3\pi }{2}$

Explanation for the correct option.

Step-1 : Finding the value of $\left({\sin }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)\right)$

Consider the given Equation as

$I=2\pi -\left({\sin }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)+{\sin }^{-1}\left(\frac{16}{65}\right)\right)$

We know that

${\sin }^{-1}x+{\sin }^{-1}y={\sin }^{-1}\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right)\left[x^2+y^2<1\right]$

Then,

$$\begin{gathered} \Rightarrow {\sin }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)={\sin }^{-1}\left(\left(\frac{4}{5}\right)\sqrt{1-{\left(\frac{5}{13}\right)}^2}+\left(\frac{5}{13}\right)\sqrt{1-{\left(\frac{4}{5}\right)}^2}\right) \\ \Rightarrow {\sin }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)={\sin }^{-1}\left(\left(\frac{4}{5}\right)\sqrt{1-\left(\frac{25}{169}\right)}+\left(\frac{5}{13}\right)\sqrt{1-\left(\frac{16}{25}\right)}\right) \\ \Rightarrow {\sin }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)={\sin }^{-1}\left(\left(\frac{4}{5}\right)\sqrt{\frac{169-25}{169}}+\left(\frac{5}{13}\right)\sqrt{\frac{25-16}{25}}\right) \\ \Rightarrow {\sin }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)={\sin }^{-1}\left(\left(\frac{4}{5}\right)\sqrt{\frac{144}{169}}+\left(\frac{5}{13}\right)\sqrt{\frac{9}{25}}\right) \\ \Rightarrow {\sin }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)={\sin }^{-1}\left(\left(\frac{4}{5}\right)\left(\frac{12}{13}\right)+\left(\frac{5}{13}\right)\left(\frac{3}{5}\right)\right) \\ \Rightarrow {\sin }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)={\sin }^{-1}\left(\left(\frac{48}{65}\right)+\left(\frac{3}{13}\right)\right) \\ \Rightarrow {\sin }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)={\sin }^{-1}\left(\frac{48+15}{65}\right) \\ \Rightarrow {\sin }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)={\sin }^{-1}\left(\frac{63}{65}\right) \end{gathered}$$

Step-2 Converting sin into cos

Consider,

$$\begin{gathered} \Rightarrow {\sin }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)=\theta \\ \Rightarrow \theta ={\sin }^{-1}\left(\frac{63}{65}\right) \\ \Rightarrow \sin \theta =\left(\frac{63}{65}\right) \end{gathered}$$

From the given Equation $\Rightarrow \cos \theta =\left(\frac{16}{65}\right)$

We know that $a^2+b^2=c^2$. Put $a=63,b=16,\text{and}c=65$

$$\begin{gathered} \Rightarrow a^2+b^2=c^2 \\ \Rightarrow {\left(63\right)}^2+{\left(16\right)}^2={\left(65\right)}^2 \\ \Rightarrow 3969+256=4225 \\ \Rightarrow 4225=4225 \end{gathered}$$

Then,

$$\begin{gathered} \Rightarrow \cos \theta =\left(\frac{16}{65}\right)=\frac{b}{h} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{16}{65}\right) \end{gathered}$$

Since,

$I=2\pi -\left[{\cos }^{-1}\left(\frac{16}{65}\right)+{\sin }^{-1}\left(\frac{16}{65}\right)\right]$

We know that

$\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\mathit{x}\mathbf{+}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{x}\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{}\mathbf{}$

Then,

$$\begin{gathered} I=2\pi -\frac{\pi }{2} \\ I=\frac{3\pi }{2} \end{gathered}$$

Hence, $2\pi -\left({\sin }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)+{\sin }^{-1}\left(\frac{16}{65}\right)\right)=\frac{3\mathrm{\pi }}{2}$

Therefore, the correct answer is option (B).