Question

The value of ${{\sin }^{-1}}\left( \dfrac{12}{13} \right)-{{\sin }^{-1}}\left( \dfrac{3}{5} \right)$ is equal to?
A. $\pi -{{\sin }^{-1}}\left( \dfrac{63}{65} \right)$
B. $\pi -{{\cos }^{-1}}\left( \dfrac{33}{65} \right)$
C. $\dfrac{\pi }{2}-{{\sin }^{-1}}\left( \dfrac{56}{65} \right)$
D. $\dfrac{\pi }{2}-{{\cos }^{-1}}\left( \dfrac{9}{65} \right)$

Answer 1

Hint: We will use some of the trigonometric identities to solve this question. We will first use the identity, $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$. Now we will replace cos B as $\sqrt{1-{{\sin }^{2}}B}$ and similarly, we will use cos A as $\sqrt{1-{{\sin }^{2}}A}$. We will take sin A as $\dfrac{12}{13}$ and sin B as $\dfrac{3}{5}$. After that we will use the identities of ${{\sin }^{-1}}x={{\cos }^{-1}}\sqrt{1-{{x}^{2}}}$ and ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$ to get the answer.

Complete step-by-step solution:
In the question, we are asked to find the value of ${{\sin }^{-1}}\left( \dfrac{12}{13} \right)-{{\sin }^{-1}}\left( \dfrac{3}{5} \right)$. So, let us consider ${{\sin }^{-1}}\left( \dfrac{12}{13} \right)$ as A and ${{\sin }^{-1}}\left( \dfrac{3}{5} \right)$ as B. To evaluate this, we will use the identity, $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$.
We know an identity that states that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, so here, we can apply it for A and B instead of $\theta $. So, in place of cos A, we will have, $\sqrt{1-{{\sin }^{2}}A}$ and in place of cos B we will have $\sqrt{1-{{\sin }^{2}}B}$. So, replacing them like this, we get the identity as follows,
$\sin \left( A-B \right)=\sin A\sqrt{1-{{\sin }^{2}}B}-\sin B\sqrt{1-{{\sin }^{2}}A}$
Now, we know that, ${{\sin }^{-1}}\left( \dfrac{12}{13} \right)=A$, so we can say that $\sin A=\dfrac{12}{13}$. Similarly, ${{\sin }^{-1}}\left( \dfrac{3}{5} \right)=B$, which can be written as, $\sin B=\dfrac{3}{5}$. We will now substitute the value of sin A and sin B in the above equation. So, we get,
 $\sin \left( A-B \right)=\dfrac{12}{13}\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}}-\dfrac{3}{5}\sqrt{1-{{\left( \dfrac{12}{13} \right)}^{2}}}$
On further simplification, we get,
$\begin{align}
  & \sin \left( A-B \right)=\dfrac{12}{13}\sqrt{1-\dfrac{9}{25}}-\dfrac{3}{5}\sqrt{1-\dfrac{144}{169}} \\
 & \Rightarrow \sin \left( A-B \right)=\dfrac{12}{13}\sqrt{\dfrac{16}{25}}-\dfrac{3}{5}\sqrt{\dfrac{25}{169}} \\
 & \Rightarrow \sin \left( A-B \right)=\dfrac{12}{13}\times \dfrac{4}{5}-\dfrac{3}{5}\times \dfrac{5}{13} \\
 & \Rightarrow \sin \left( A-B \right)=\dfrac{48}{65}-\dfrac{15}{65} \\
 & \Rightarrow \sin \left( A-B \right)=\dfrac{33}{65} \\
\end{align}$
So, we get the value of $A-B={{\sin }^{-1}}\left( \dfrac{33}{65} \right)$ . This can be written as follows also,
${{\sin }^{-1}}\left( \dfrac{12}{13} \right)-{{\sin }^{-1}}\left( \dfrac{3}{5} \right)={{\sin }^{-1}}\left( \dfrac{33}{65} \right)$
We know that ${{\sin }^{-1}}x={{\cos }^{-1}}\sqrt{1-{{x}^{2}}}$ and we will substitute x with $\dfrac{33}{65}$. So, we get,
$\begin{align}
  & {{\sin }^{-1}}\left( \dfrac{33}{65} \right)={{\cos }^{-1}}\sqrt{1-{{\left( \dfrac{33}{65} \right)}^{2}}} \\
 & \Rightarrow {{\sin }^{-1}}\left( \dfrac{33}{65} \right)={{\cos }^{-1}}\sqrt{1-\dfrac{1089}{4225}} \\
 & \Rightarrow {{\sin }^{-1}}\left( \dfrac{33}{65} \right)={{\cos }^{-1}}\sqrt{\dfrac{3136}{4225}} \\
 & \Rightarrow {{\sin }^{-1}}\left( \dfrac{33}{65} \right)={{\cos }^{-1}}\left( \dfrac{56}{65} \right) \\
\end{align}$
Now, we will apply the identity, ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$, which can be written as, ${{\cos }^{-1}}x=\dfrac{\pi }{2}-{{\sin }^{-1}}x$. Now, we will take the value of x as $\dfrac{56}{65}$. So, we get,
${{\cos }^{-1}}\left( \dfrac{56}{65} \right)=\dfrac{\pi }{2}-{{\sin }^{-1}}\left( \dfrac{56}{65} \right)$
Hence, we get the value of ${{\sin }^{-1}}\left( \dfrac{12}{13} \right)-{{\sin }^{-1}}\left( \dfrac{3}{5} \right)={{\sin }^{-1}}\left( \dfrac{33}{65} \right)$ as,
${{\sin }^{-1}}\left( \dfrac{12}{13} \right)-{{\sin }^{-1}}\left( \dfrac{3}{5} \right)=\dfrac{\pi }{2}-{{\sin }^{-1}}\left( \dfrac{56}{65} \right)$ .
Therefore, the correct answer is option C.

Note: The students can use a short cut identity in place of the long method that we have used in the solution. They can use the identity, ${{\sin }^{-1}}x-{{\sin }^{-1}}y={{\sin }^{-1}}\left( \sqrt{1-{{y}^{2}}}\times x-\sqrt{1-{{x}^{2}}}\times y \right)$ and after this they can apply the identity of, ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$.