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Question: The value of \({{\sin }^{-1}}\left( \dfrac{12}{13} \right)-{{\sin }^{-1}}\left( \dfrac{3}{5} \right)...

The value of sin⁡−1(1213)−sin⁡−1(35){{\sin }^{-1}}\left( \dfrac{12}{13} \right)-{{\sin }^{-1}}\left( \dfrac{3}{5} \right) is equal to?
A. π−sin⁡−1(6365)\pi -{{\sin }^{-1}}\left( \dfrac{63}{65} \right)
B. π−cos⁡−1(3365)\pi -{{\cos }^{-1}}\left( \dfrac{33}{65} \right)
C. π2−sin⁡−1(5665)\dfrac{\pi }{2}-{{\sin }^{-1}}\left( \dfrac{56}{65} \right)
D. π2−cos⁡−1(965)\dfrac{\pi }{2}-{{\cos }^{-1}}\left( \dfrac{9}{65} \right)

Explanation

Solution

We will use some of the trigonometric identities to solve this question. We will first use the identity, sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B. Now we will replace cos B as 1−sin⁡2B\sqrt{1-{{\sin }^{2}}B} and similarly, we will use cos A as 1−sin⁡2A\sqrt{1-{{\sin }^{2}}A}. We will take sin A as 1213\dfrac{12}{13} and sin B as 35\dfrac{3}{5}. After that we will use the identities of sin⁡−1x=cos⁡−11−x2{{\sin }^{-1}}x={{\cos }^{-1}}\sqrt{1-{{x}^{2}}} and sin⁡−1x+cos⁡−1x=π2{{\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(1213)−sin⁡−1(35){{\sin }^{-1}}\left( \dfrac{12}{13} \right)-{{\sin }^{-1}}\left( \dfrac{3}{5} \right). So, let us consider sin⁡−1(1213){{\sin }^{-1}}\left( \dfrac{12}{13} \right) as A and sin⁡−1(35){{\sin }^{-1}}\left( \dfrac{3}{5} \right) as B. To evaluate this, we will use the identity, sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B.
We know an identity that states that sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1, so here, we can apply it for A and B instead of θ\theta . So, in place of cos A, we will have, 1−sin⁡2A\sqrt{1-{{\sin }^{2}}A} and in place of cos B we will have 1−sin⁡2B\sqrt{1-{{\sin }^{2}}B}. So, replacing them like this, we get the identity as follows,
sin⁡(A−B)=sin⁡A1−sin⁡2B−sin⁡B1−sin⁡2A\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(1213)=A{{\sin }^{-1}}\left( \dfrac{12}{13} \right)=A, so we can say that sin⁡A=1213\sin A=\dfrac{12}{13}. Similarly, sin⁡−1(35)=B{{\sin }^{-1}}\left( \dfrac{3}{5} \right)=B, which can be written as, sin⁡B=35\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⁡(A−B)=12131−(35)2−351−(1213)2\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,
sin⁡(A−B)=12131−925−351−144169 ⇒sin⁡(A−B)=12131625−3525169 ⇒sin⁡(A−B)=1213×45−35×513 ⇒sin⁡(A−B)=4865−1565 ⇒sin⁡(A−B)=3365 \begin{aligned} & \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{aligned}
So, we get the value of A−B=sin⁡−1(3365)A-B={{\sin }^{-1}}\left( \dfrac{33}{65} \right) . This can be written as follows also,
sin⁡−1(1213)−sin⁡−1(35)=sin⁡−1(3365){{\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⁡−1x=cos⁡−11−x2{{\sin }^{-1}}x={{\cos }^{-1}}\sqrt{1-{{x}^{2}}} and we will substitute x with 3365\dfrac{33}{65}. So, we get,
sin⁡−1(3365)=cos⁡−11−(3365)2 ⇒sin⁡−1(3365)=cos⁡−11−10894225 ⇒sin⁡−1(3365)=cos⁡−131364225 ⇒sin⁡−1(3365)=cos⁡−1(5665) \begin{aligned} & {{\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{aligned}
Now, we will apply the identity, sin⁡−1x+cos⁡−1x=π2{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}, which can be written as, cos⁡−1x=π2−sin⁡−1x{{\cos }^{-1}}x=\dfrac{\pi }{2}-{{\sin }^{-1}}x. Now, we will take the value of x as 5665\dfrac{56}{65}. So, we get,
cos⁡−1(5665)=π2−sin⁡−1(5665){{\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(1213)−sin⁡−1(35)=sin⁡−1(3365){{\sin }^{-1}}\left( \dfrac{12}{13} \right)-{{\sin }^{-1}}\left( \dfrac{3}{5} \right)={{\sin }^{-1}}\left( \dfrac{33}{65} \right) as,
sin⁡−1(1213)−sin⁡−1(35)=π2−sin⁡−1(5665){{\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⁡−1x−sin⁡−1y=sin⁡−1(1−y2×x−1−x2×y){{\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⁡−1x+cos⁡−1x=π2{{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}.