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Question: If \( {\operatorname{Sin} ^{ - 1}}\left( x \right) + {\operatorname{Sin} ^{ - 1}}\left( y \right) = ...

If Sin⁡−1(x)+Sin⁡−1(y)=2π3{\operatorname{Sin} ^{ - 1}}\left( x \right) + {\operatorname{Sin} ^{ - 1}}\left( y \right) = \dfrac{{2\pi }}{3} , then what is the value of Cos⁡−1(x)+Cos⁡−1(y){\operatorname{Cos} ^{ - 1}}\left( x \right) + {\operatorname{Cos} ^{ - 1}}\left( y \right) ?
A) 2π3\dfrac{{2\pi }}{3}
B) π3\dfrac{\pi }{3}
C) π6\dfrac{\pi }{6}
D) π\pi

Explanation

Solution

Hint : We know that the above equations are given in the form of inverse trigonometric functions or anti-trigonometric functions. Inverse trigonometric functions are the inverse functions of the trigonometric functions. In Inverse trigonometric functions there is an identity called inverse sum identity, we can solve this question by using Sin⁡−1(x)+Cos⁡−1(x)=π2{\operatorname{Sin} ^{ - 1}}\left( x \right) + {\operatorname{Cos} ^{ - 1}}\left( x \right) = \dfrac{\pi }{2} identity.
Formula: Inverse sum identity- Sin⁡−1(x)+Cos⁡−1(x)=π2{\operatorname{Sin} ^{ - 1}}\left( x \right) + {\operatorname{Cos} ^{ - 1}}\left( x \right) = \dfrac{\pi }{2} , for all x∈[−1,1]x \in \left[ { - 1,1} \right]

Complete step-by-step answer :
We have,
Sin⁡−1(x)+Sin⁡−1(y)=2π3{\operatorname{Sin} ^{ - 1}}\left( x \right) + {\operatorname{Sin} ^{ - 1}}\left( y \right) = \dfrac{{2\pi }}{3} … (i)\left( i \right)
We know that, Sin⁡−1(x)+Cos⁡−1(x)=π2{\operatorname{Sin} ^{ - 1}}\left( x \right) + {\operatorname{Cos} ^{ - 1}}\left( x \right) = \dfrac{\pi }{2} for all x∈[−1,1]x \in \left[ { - 1,1} \right] [Inverse sum identity]
Now, we will shift Cos⁡−1(x){\operatorname{Cos} ^{ - 1}}\left( x \right) to Right hand side:
Sin⁡−1(x)=π2−Cos⁡−1(x){\operatorname{Sin} ^{ - 1}}\left( x \right) = \dfrac{\pi }{2} - {\operatorname{Cos} ^{ - 1}}\left( x \right)
If we replace xx by yy in the inverse sum identity. Then, we get
Sin⁡−1(y)+Cos⁡−1(y)=π2{\operatorname{Sin} ^{ - 1}}\left( y \right) + {\operatorname{Cos} ^{ - 1}}\left( y \right) = \dfrac{\pi }{2}
Now, we will Shift Cos⁡−1(y){\operatorname{Cos} ^{ - 1}}\left( y \right) to Right hand side:
Sin⁡−1(y)=π2−Cos⁡−1(y){\operatorname{Sin} ^{ - 1}}\left( y \right) = \dfrac{\pi }{2} - {\operatorname{Cos} ^{ - 1}}\left( y \right)
Now, we will substitute the value of Sin⁡−1(x){\operatorname{Sin} ^{ - 1}}\left( x \right) and Sin⁡−1(y){\operatorname{Sin} ^{ - 1}}\left( y \right) in the equation (i)\left( i \right)
⇒π2−Cos⁡−1(x)+π2−Cos⁡−1(y)=2π3\Rightarrow \dfrac{\pi }{2} - {\operatorname{Cos} ^{ - 1}}\left( x \right) + \dfrac{\pi }{2} - {\operatorname{Cos} ^{ - 1}}\left( y \right) = \dfrac{{2\pi }}{3}
⇒π2+π2−Cos⁡−1(x)−Cos⁡−1(y)=2π3\Rightarrow \dfrac{\pi }{2} + \dfrac{\pi }{2} - {\operatorname{Cos} ^{ - 1}}\left( x \right) - {\operatorname{Cos} ^{ - 1}}\left( y \right) = \dfrac{{2\pi }}{3}
Now, we will take the LCM of π2+π2\dfrac{\pi }{2} + \dfrac{\pi }{2}
⇒π−Cos⁡−1(x)−Cos⁡−1(y)=2π3\Rightarrow \pi - {\operatorname{Cos} ^{ - 1}}\left( x \right) - {\operatorname{Cos} ^{ - 1}}\left( y \right) = \dfrac{{2\pi }}{3}
Shift 2π3\dfrac{{2\pi }}{3} to L.H.S. and −Cos⁡−1(x)- {\operatorname{Cos} ^{ - 1}}\left( x \right) , −Cos⁡−1(y)- {\operatorname{Cos} ^{ - 1}}\left( y \right) to R.H.S.
⇒π−2π3=Cos⁡−1(x)+Cos⁡−1(y)\Rightarrow \pi - \dfrac{{2\pi }}{3} = {\operatorname{Cos} ^{ - 1}}\left( x \right) + {\operatorname{Cos} ^{ - 1}}\left( y \right)
We can also write it as:
⇒Cos⁡−1(x)+Cos⁡−1(y)=π−2π3\Rightarrow {\operatorname{Cos} ^{ - 1}}\left( x \right) + {\operatorname{Cos} ^{ - 1}}\left( y \right) = \pi - \dfrac{{2\pi }}{3}
Now, we will take the LCM of π−2π3\pi - \dfrac{{2\pi }}{3}
⇒Cos⁡−1(x)+Cos⁡−1(y)=3π−2π3\Rightarrow {\operatorname{Cos} ^{ - 1}}\left( x \right) + {\operatorname{Cos} ^{ - 1}}\left( y \right) = \dfrac{{3\pi - 2\pi }}{3}
∴Cos⁡−1(x)+Cos⁡−1(y)=π3\therefore {\operatorname{Cos} ^{ - 1}}\left( x \right) + {\operatorname{Cos} ^{ - 1}}\left( y \right) = \dfrac{\pi }{3}
Hence, the correct option is 2.
So, the correct answer is “Option 2”.

Note : There are many formulae/identities in inverse trigonometric functions to solve different types of questions, so choose the identity carefully. For example: The identity Sin⁡−1(x)−Sin⁡−1(y)=π{\operatorname{Sin} ^{ - 1}}\left( x \right) - {\operatorname{Sin} ^{ - 1}}\left( y \right) = \pi looks very much similar to the given equation Sin⁡−1(x)+Sin⁡−1(y)=2π3{\operatorname{Sin} ^{ - 1}}\left( x \right) + {\operatorname{Sin} ^{ - 1}}\left( y \right) = \dfrac{{2\pi }}{3} but if we observe carefully we can see that in Sin⁡−1(x)−Sin⁡−1(y)=π{\operatorname{Sin} ^{ - 1}}\left( x \right) - {\operatorname{Sin} ^{ - 1}}\left( y \right) = \pi identity, there is subtraction between two inverse trigonometric functions whereas there is addition between the two inverse trigonometric functions which we need to solve. We know that there are different identities to solve different types of inverse trigonometric functions:
Sin⁡−1(x)+Cos⁡−1(x)=π2,x∈[−1,1]{\operatorname{Sin} ^{ - 1}}\left( x \right) + {\operatorname{Cos} ^{ - 1}}\left( x \right) = \dfrac{\pi }{2},x \in \left[ { - 1,1} \right]
tan⁡−1(x)+cot⁡−1(x)=π2,x∈R{\tan ^{ - 1}}\left( x \right) + {\cot ^{ - 1}}\left( x \right) = \dfrac{\pi }{2},x \in R
cos⁡ec−1(x)+sec⁡−1(x)=π2,∣x∣⩾1\cos e{c^{ - 1}}\left( x \right) + {\sec ^{ - 1}}\left( x \right) = \dfrac{\pi }{2},\left| x \right| \geqslant 1
We use them according to the given problem.