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Question: If \[{{\sin }^{-1}}(x)=\theta +\beta \] and \[{{\sin }^{-1}}(y)=\theta -\beta \] , then \[1+xy=\] ...

If sin⁡−1(x)=θ+β{{\sin }^{-1}}(x)=\theta +\beta and sin⁡−1(y)=θ−β{{\sin }^{-1}}(y)=\theta -\beta , then 1+xy=1+xy=
A.sin⁡2θ+sin⁡2β{{\sin }^{2}}\theta +{{\sin }^{2}}\beta
B.sin⁡2θ+cos⁡2β{{\sin }^{2}}\theta +{{\cos }^{2}}\beta
C.cos⁡2θ+cos⁡2β{{\cos }^{2}}\theta +{{\cos }^{2}}\beta
D.cos⁡2θ+sin⁡2β{{\cos }^{2}}\theta +{{\sin }^{2}}\beta

Explanation

Solution

Hint : Inverse trigonometric functions are simply defined as the inverse functions of the basic trigonometric functions which are sine, cosine, tangent, cotangent, secant, and cosecant functions. Sine can be defined as the ratio of the opposite side of a right triangle to its hypotenuse. The sine function has values positive or negative upon the quadrants. Sine is positive in the first and second quadrants while it is negative for the third and fourth quadrants.
They are also termed as arcus functions (also called as anti trigonometric functions or cyclometric functions). The inverse of these six trigonometric functions are:
Arcsine, Arccosine, Arctangent, Arcsecant, Arccosecant, Arccotangent.

Complete step-by-step answer :
Now, according to the question:
As we are given that:
sin⁡−1(x)=θ+β{{\sin }^{-1}}(x)=\theta +\beta
⇒x=sin⁡(θ+β)\Rightarrow x=\sin (\theta +\beta )
And sin⁡−1(y)=θ−β{{\sin }^{-1}}(y)=\theta -\beta
⇒y=sin⁡(θ−β)\Rightarrow y=\sin (\theta -\beta )
As, according to given in question: 1+xy1+xy
Now, substituting values in this we will get:
⇒1+sin⁡(θ+β)sin⁡(θ−β)\Rightarrow 1+\sin (\theta +\beta )\sin (\theta -\beta )
Now, we will use the formulas such that:
sin⁡(A−B)=sin⁡Acos⁡B−sin⁡Bcos⁡A\sin (A-B)=\sin A\cos B-\sin B\cos A and sin⁡(A+B)=sin⁡Acos⁡B+sin⁡Bcos⁡A\sin (A+B)=\sin A\cos B+\sin B\cos A
⇒1+[(sin⁡θcos⁡β+cos⁡θsin⁡β)(sin⁡θcos⁡β−cos⁡θsin⁡β)]\Rightarrow 1+\left[ (\sin \theta \cos \beta +\cos \theta \sin \beta )(\sin \theta \cos \beta -\cos \theta \sin \beta ) \right]
⇒1+[(sin⁡θcos⁡β)2−(cos⁡θsin⁡β)2]\Rightarrow 1+\left[ {{(\sin \theta \cos \beta )}^{2}}-{{(\cos \theta \sin \beta )}^{2}} \right]
⇒sin⁡2θ+cos⁡2θ+[(sin⁡θcos⁡β)2−(cos⁡θsin⁡β)2]\Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta +\left[ {{(\sin \theta \cos \beta )}^{2}}-{{(\cos \theta \sin \beta )}^{2}} \right]
⇒sin⁡2θ[1+cos⁡2β]+cos⁡2θ[1−sin⁡2β]\Rightarrow {{\sin }^{2}}\theta [1+{{\cos }^{2}}\beta ]+{{\cos }^{2}}\theta [1-{{\sin }^{2}}\beta ]
⇒sin⁡2θ+sin⁡2θcos⁡2β+cos⁡2θcos⁡2β\Rightarrow {{\sin }^{2}}\theta +{{\sin }^{2}}\theta {{\cos }^{2}}\beta +{{\cos }^{2}}\theta {{\cos }^{2}}\beta
⇒sin⁡2θ+cos⁡2β[sin⁡2θ+cos⁡2θ]\Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\beta [{{\sin }^{2}}\theta +{{\cos }^{2}}\theta ]
⇒sin⁡2θ+cos⁡2β\Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\beta
Hence, 1+xy=sin⁡2θ+cos⁡2β1+xy={{\sin }^{2}}\theta +{{\cos }^{2}}\beta
Therefore, from all the above options BB is the correct option.
So, the correct answer is “Option B”.

Note : One example of an inverse trigonometric function is the angle of depression and angle of elevation. An example of people using inverse trigonometric functions would be builders such as construction workers, architects, in the creation of bike ramps and many more.