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Question: If \[{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \pi \] , prove that \[x\sqrt {1 - {x^2}} ...

If sin⁡−1x+sin⁡−1y+sin⁡−1z=π{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \pi , prove that x1−x2+y1−y2+z1−z2=2xyzx\sqrt {1 - {x^2}} + y\sqrt {1 - {y^2}} + z\sqrt {1 - {z^2}} = 2xyz .

Explanation

Solution

Hint : We are asked to prove the expression x1−x2+y1−y2+z1−z2=2xyzx\sqrt {1 - {x^2}} + y\sqrt {1 - {y^2}} + z\sqrt {1 - {z^2}} = 2xyz . And one more expression is given sin⁡−1x+sin⁡−1y+sin⁡−1z=π{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \pi , use this expression to find the value of xx , yy and zz . Use these values to prove the required expression.

Complete step-by-step answer :
Given, sin⁡−1x+sin⁡−1y+sin⁡−1z=π{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \pi (i)
And we are asked to prove x1−x2+y1−y2+z1−z2=2xyzx\sqrt {1 - {x^2}} + y\sqrt {1 - {y^2}} + z\sqrt {1 - {z^2}} = 2xyz (ii)
Let
sin⁡−1x=A{\sin ^{ - 1}}x = A (iii)
sin⁡−1y=B{\sin ^{ - 1}}y = B (iv)
sin⁡−1z=C{\sin ^{ - 1}}z = C (v)
Now, we use equations (iii), (iv) and (v) in equation (i), we get
A+B+C=πA + B + C = \pi (vi)
Now,
sin⁡−1x=A{\sin ^{ - 1}}x = A
⇒x=sin⁡A\Rightarrow x = \sin A (vii)
sin⁡−1y=B{\sin ^{ - 1}}y = B
⇒y=sin⁡B\Rightarrow y = \sin B (viii)
sin⁡−1z=C{\sin ^{ - 1}}z = C
⇒z=sin⁡C\Rightarrow z = \sin C (ix)
Now using equations (vii), (viii) and (ix) in L.H.S of equation (ii), we get
L.H.S=sin⁡A1−sin⁡2A+sin⁡B1−sin⁡2B+sin⁡C1−sin⁡2CL.H.S = \sin A\sqrt {1 - {{\sin }^2}A} + \sin B\sqrt {1 - {{\sin }^2}B} + \sin C\sqrt {1 - {{\sin }^2}C}

=sin⁡Acos⁡A+sin⁡Bcos⁡B+sin⁡Ccos⁡C =2sin⁡Acos⁡A+2sin⁡Bcos⁡B2+sin⁡Ccos⁡C  = \sin A\cos A + \sin B\cos B + \sin C\cos C \\\ = \dfrac{{2\sin A\cos A + 2\sin B\cos B}}{2} + \sin C\cos C \\\

Using the trigonometric identity 2sin⁡xcos⁡x=sin⁡2x2\sin x\cos x = \sin 2x in the above expression we get,
L.H.S=sin⁡2A+sin⁡2B2+sin⁡Ccos⁡CL.H.S = \dfrac{{\sin 2A + \sin 2B}}{2} + \sin C\cos C
Now, using the trigonometric identity sin⁡X+sin⁡Y=2sin⁡(X+Y2)cos⁡(X−Y2)\sin X + \sin Y = 2\sin \left( {\dfrac{{X + Y}}{2}} \right)\cos \left( {\dfrac{{X - Y}}{2}} \right) in the above equation we get,
L.H.S=2sin⁡(2A+2B2)cos⁡(2A−2B2)2+sin⁡Ccos⁡CL.H.S = \dfrac{{2\sin \left( {\dfrac{{2A + 2B}}{2}} \right)\cos \left( {\dfrac{{2A - 2B}}{2}} \right)}}{2} + \sin C\cos C
=sin⁡(A+B)cos⁡(A−B)+sin⁡Ccos⁡C= \sin (A + B)\cos (A - B) + \sin C\cos C
Now writing A+B=π−CA + B = \pi - C and C=π−(A+B)C = \pi - \left( {A + B} \right) in the above expression from equation (vi), we get
L.H.S=sin⁡(π−C)cos⁡(A−B)+sin⁡Ccos⁡(π−(A+B))L.H.S = \sin (\pi - C)\cos (A - B) + \sin C\cos \left( {\pi - \left( {A + B} \right)} \right)
=sin⁡Ccos⁡(A−B)−sin⁡Ccos⁡(A+B)= \sin C\cos (A - B) - \sin C\cos (A + B)
=sin⁡C(cos⁡(A−B)−cos⁡(A+B))= \sin C\left( {\cos (A - B) - \cos (A + B)} \right)
The term cos⁡(A−B)−cos⁡(A+B)\cos (A - B) - \cos (A + B) can be written as 2sin⁡Asin⁡B2\sin A\sin B .
∴L.H.S=sin⁡C(2sin⁡Asin⁡B)\therefore L.H.S = \sin C\left( {2\sin A\sin B} \right)
=2sin⁡Asin⁡Bsin⁡C= 2\sin A\sin B\sin C (x)
Using (vii), (viii) and (ix) in equation (x), we get
L.H.S=2xyzL.H.S = 2xyz
Therefore,
x1−x2+y1−y2+z1−z2=2xyzx\sqrt {1 - {x^2}} + y\sqrt {1 - {y^2}} + z\sqrt {1 - {z^2}} = 2xyz
So, the correct answer is “2xyz”.

Note : In such types of questions, where we need to prove an expression, always start from one side of the equation either the left hand side or right hand side. For solving the problem in an easy way go for the side which can be further simplified. Here, we started from the left hand side, as we can observe that this side can be further simplified while the right hand side is already in simplified form. Similarly proceed for such types of questions and look for the hints or conditions given and use them to equate both the sides. Also, for questions involving trigonometric terms remember the trigonometric identities as it will help you to simplify.