Question

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

Answer 1

Hint: We are asked to prove the expression \[x\sqrt {1 - {x^2}} + y\sqrt {1 - {y^2}} + z\sqrt {1 - {z^2}} = 2xyz\] . And one more expression is given \[{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \pi \] , use this expression to find the value of \[x\] , \[y\] and \[z\] . Use these values to prove the required expression.

Complete step-by-step answer:
Given, \[{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \pi \] (i)
And we are asked to prove \[x\sqrt {1 - {x^2}} + y\sqrt {1 - {y^2}} + z\sqrt {1 - {z^2}} = 2xyz\] (ii)
Let
 \[{\sin ^{ - 1}}x = A\] (iii)
 \[{\sin ^{ - 1}}y = B\] (iv)
 \[{\sin ^{ - 1}}z = C\] (v)
Now, we use equations (iii), (iv) and (v) in equation (i), we get
 \[A + B + C = \pi \] (vi)
Now,
 \[{\sin ^{ - 1}}x = A\]
 \[ \Rightarrow x = \sin A\] (vii)
 \[{\sin ^{ - 1}}y = B\]
 \[ \Rightarrow y = \sin B\] (viii)
 \[{\sin ^{ - 1}}z = C\]
 \[ \Rightarrow z = \sin C\] (ix)
Now using equations (vii), (viii) and (ix) in L.H.S of equation (ii), we get
 \[L.H.S = \sin A\sqrt {1 - {{\sin }^2}A} + \sin B\sqrt {1 - {{\sin }^2}B} + \sin C\sqrt {1 - {{\sin }^2}C} \]
 \[
   = \sin A\cos A + \sin B\cos B + \sin C\cos C \\
   = \dfrac{{2\sin A\cos A + 2\sin B\cos B}}{2} + \sin C\cos C \\
\]
Using the trigonometric identity \[2\sin x\cos x = \sin 2x\] in the above expression we get,
 \[L.H.S = \dfrac{{\sin 2A + \sin 2B}}{2} + \sin C\cos C\]
Now, using the trigonometric identity \[\sin X + \sin Y = 2\sin \left( {\dfrac{{X + Y}}{2}} \right)\cos \left( {\dfrac{{X - Y}}{2}} \right)\] in the above equation we get,
 \[L.H.S = \dfrac{{2\sin \left( {\dfrac{{2A + 2B}}{2}} \right)\cos \left( {\dfrac{{2A - 2B}}{2}} \right)}}{2} + \sin C\cos C\]
 \[ = \sin (A + B)\cos (A - B) + \sin C\cos C\]
Now writing \[A + B = \pi - C\] and \[C = \pi - \left( {A + B} \right)\] in the above expression from equation (vi), we get
 \[L.H.S = \sin (\pi - C)\cos (A - B) + \sin C\cos \left( {\pi - \left( {A + B} \right)} \right)\]
 \[ = \sin C\cos (A - B) - \sin C\cos (A + B)\]
 \[ = \sin C\left( {\cos (A - B) - \cos (A + B)} \right)\]
The term \[\cos (A - B) - \cos (A + B)\] can be written as \[2\sin A\sin B\] .
 \[\therefore L.H.S = \sin C\left( {2\sin A\sin B} \right)\]
 \[ = 2\sin A\sin B\sin C\] (x)
Using (vii), (viii) and (ix) in equation (x), we get
 \[L.H.S = 2xyz\]
Therefore,
 \[x\sqrt {1 - {x^2}} + y\sqrt {1 - {y^2}} + z\sqrt {1 - {z^2}} = 2xyz\]
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.