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Question: Point P(x,y) satisfying the equation \[{{\operatorname{Sin}}^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}\l...

Point P(x,y) satisfying the equation Sin⁡−1x+cos⁡−1y+cos⁡−1(2xy)=π2{{\operatorname{Sin}}^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}\left( 2xy \right)=\dfrac{\pi }{2} lies on?
1)1)the bisector of the first and third quadrant
2)2)bisector of the second and fourth quadrant
3)3)the rectangle formed by the lines x=±1&y=±1x=\pm 1\And y=\pm 1
4)4)a unit circle with centre at the origin

Explanation

Solution

To solve these types of questions of inverse trigonometric ratios, we should know some of the properties of trigonometric as well as inverse trigonometric functions. The properties that we should know are as follows: T(T−1x)=xT\left( {{T}^{-1}}x \right)=x here T is a trigonometric ratio like sine, cosine, tangent, etc. Also, we should know that sin⁡(cos⁡−1x)=1−x2\sin \left( {{\cos }^{-1}}x \right)=\sqrt{1-{{x}^{2}}}. The trigonometric property we will use states that cos⁡(π2−A)=sin⁡A\cos \left( \dfrac{\pi }{2}-A \right)=\sin A, and the other is expansion of difference of angle for cosine cos⁡(a−b)=cos⁡acos⁡b+sin⁡asin⁡b\cos \left( a-b \right)=\cos a\cos b+\sin a\sin b. We will use these properties to simplify the given equation and find the correct option for it.

Complete step-by-step solution:
We are given the equation as Sin⁡−1x+cos⁡−1y+cos⁡−1(2xy)=π2{{\operatorname{Sin}}^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}\left( 2xy \right)=\dfrac{\pi }{2}. Simplifying this equation, we can write it as cos⁡−1y+cos⁡−1(2xy)=π2−sin⁡−1x{{\cos }^{-1}}y+{{\cos }^{-1}}\left( 2xy \right)=\dfrac{\pi }{2}-{{\sin }^{-1}}x. Taking cosine of both sides of the above equation, we get
⇒cos⁡(cos⁡−1y+cos⁡−1(2xy))=cos⁡(π2−sin⁡−1x)\Rightarrow \cos \left( {{\cos }^{-1}}y+{{\cos }^{-1}}\left( 2xy \right) \right)=\cos \left( \dfrac{\pi }{2}-{{\sin }^{-1}}x \right)
Using the sum of angles for cosine, we can simplify the left side of the above equation as follows. Then, we will use the property of inverse trigonometric ratios which states sin⁡(cos⁡−1x)=1−x2\sin \left( {{\cos }^{-1}}x \right)=\sqrt{1-{{x}^{2}}} to further simplify the equation as
⇒y×2xy−1−y2×1−(2xy)2=cos⁡(π2−sin⁡−1x)\Rightarrow y\times 2xy-\sqrt{1-{{y}^{2}}}\times \sqrt{1-{{\left( 2xy \right)}^{2}}}=\cos \left( \dfrac{\pi }{2}-{{\sin }^{-1}}x \right)
Using the property cos⁡(π2−A)=sin⁡A\cos \left( \dfrac{\pi }{2}-A \right)=\sin A, we can simplify the right side of the equation as follows 2xy2−1−y21−(2xy)2=x2x{{y}^{2}}-\sqrt{1-{{y}^{2}}}\sqrt{1-{{\left( 2xy \right)}^{2}}}=x.
Taking the term 2xy22x{{y}^{2}} to the right side, and taking the square of both sides of the equation. We get
⇒(−1−y21−(2xy)2)2=(x−2xy2)2\Rightarrow {{\left( -\sqrt{1-{{y}^{2}}}\sqrt{1-{{\left( 2xy \right)}^{2}}} \right)}^{2}}={{\left( x-2x{{y}^{2}} \right)}^{2}}
Simplifying the above equation, we can write it as
⇒(1−y2)(1−(2xy)2)=x2−4x2y2+4x2y4\Rightarrow \left( 1-{{y}^{2}} \right)\left( 1-{{\left( 2xy \right)}^{2}} \right)={{x}^{2}}-4{{x}^{2}}{{y}^{2}}+4{{x}^{2}}{{y}^{4}}
⇒1−y2−4x2y2+4x2y4=x2−4x2y2+4x2y4\Rightarrow 1-{{y}^{2}}-4{{x}^{2}}{{y}^{2}}+4{{x}^{2}}{{y}^{4}}={{x}^{2}}-4{{x}^{2}}{{y}^{2}}+4{{x}^{2}}{{y}^{4}}

& \Rightarrow 1-{{y}^{2}}={{x}^{2}} \\\ & \Rightarrow {{x}^{2}}+{{y}^{2}}=1 \\\ \end{aligned}$$ This is the equation of a circle with centre at origin and radius as one. **Hence, the correct option is D.** **Note:** To solve these types of questions, we should know all properties of inverse trigonometric and trigonometric functions. After simplifying the equation using these properties, identify the type of the equation. That is the conic that the equation represents. For this question, the conic was a circle with a centre at the origin. Calculation mistakes while solving these questions should be avoided.