if sin 1x sin 1y sin 1z pi then x 4 y 4 z 4 4x 2y 2z 2

If $${\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \pi ,$$ then $${x^4} + {y^4} + {z^4} + 4{x^2}{y^2}{z^2} = k\left( {{x^2}{y^2} + {y^2}{z^2} + {z^2}{x^2}} \right).$$ where $$k =$$

A. 1

B. 2

C. 4

D. None of these

Answer : 2

Solution :
$$\eqalign{ & {\text{We have, }}{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \pi - {\sin ^{ - 1}}z \cr & {\text{or, }}x\sqrt {\left( {1 - {y^2}} \right)} + y\sqrt {\left( {1 - {x^2}} \right)} = z \cr & {\text{or, }}{x^2}\left( {1 - {y^2}} \right) = {z^2} + {y^2}\left( {1 - {x^2}} \right) - 2yz\sqrt {\left( {1 - {x^2}} \right)} \cr & {\text{or, }}{\left( {{x^2} - {z^2} - {y^2}} \right)^2} = 4{y^2}{z^2}\left( {1 - {x^2}} \right) \cr & {\text{or, }}{x^4} + {y^2} + {z^4} - 2{x^2}{z^2} + 2{y^2}{z^2} - 2{x^2}{y^2} + 4{x^2}{y^2}{z^2} - 4{y^2}{z^2} = 0 \cr & {\text{or, }}{x^4} + {y^4} + {z^4} + 4{x^2}{y^2}{z^2} = 2\left( {{x^2}{y^2} + {y^2}{z^2} + {z^2}{x^2}} \right) \cr & \therefore k = 2 \cr} $$

Releted MCQ Question on
Trigonometry >> Inverse Trigonometry Function

Releted Question 1

The value of $$\tan \left[ {{{\cos }^{ - 1}}\left( {\frac{4}{5}} \right) + {{\tan }^{ - 1}}\left( {\frac{2}{3}} \right)} \right]$$ is

A. $$\frac{6}{{17}}$$

B. $$\frac{7}{{16}}$$

C. $$\frac{16}{{7}}$$

D. none

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Releted Question 2

If we consider only the principle values of the inverse trigonometric functions then the value of $$\tan \left( {{{\cos }^{ - 1}}\frac{1}{{5\sqrt 2 }} - {{\sin }^{ - 1}}\frac{4}{{\sqrt {17} }}} \right)$$ is

A. $$\frac{{\sqrt {29} }}{3}$$

B. $$\frac{{29}}{3}$$

C. $$\frac{{\sqrt {3}}}{29}$$

D. $$\frac{{3}}{29}$$

View Answer

Releted Question 3

The number of real solutions of $${\tan ^{ - 1}}\sqrt {x\left( {x + 1} \right)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = \frac{\pi }{2}$$ is

A. zero

B. one

C. two

D. infinite

View Answer

Releted Question 4

If $${\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - .....} \right) + {\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - .....} \right) = \frac{\pi }{2}$$ for $$0 < \left| x \right| < \sqrt 2 ,$$ then $$x$$ equals

A. $$ \frac{1}{2}$$

B. 1

C. $$ - \frac{1}{2}$$

D. $$- 1$$

View Answer

If $${\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \pi ,$$ then $${x^4} + {y^4} + {z^4} + 4{x^2}{y^2}{z^2} = k\left( {{x^2}{y^2} + {y^2}{z^2} + {z^2}{x^2}} \right).$$ where $$k =$$

Releted MCQ Question on
Trigonometry >> Inverse Trigonometry Function

The value of $$\tan \left[ {{{\cos }^{ - 1}}\left( {\frac{4}{5}} \right) + {{\tan }^{ - 1}}\left( {\frac{2}{3}} \right)} \right]$$ is

If we consider only the principle values of the inverse trigonometric functions then the value of $$\tan \left( {{{\cos }^{ - 1}}\frac{1}{{5\sqrt 2 }} - {{\sin }^{ - 1}}\frac{4}{{\sqrt {17} }}} \right)$$ is

The number of real solutions of $${\tan ^{ - 1}}\sqrt {x\left( {x + 1} \right)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = \frac{\pi }{2}$$ is

If $${\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - .....} \right) + {\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - .....} \right) = \frac{\pi }{2}$$ for $$0 < \left| x \right| < \sqrt 2 ,$$ then $$x$$ equals

Practice More Releted MCQ Question on
Inverse Trigonometry Function

Practice More MCQ Question on Maths Section

If $${\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \pi ,$$ then $${x^4} + {y^4} + {z^4} + 4{x^2}{y^2}{z^2} = k\left( {{x^2}{y^2} + {y^2}{z^2} + {z^2}{x^2}} \right).$$ where $$k =$$

Releted MCQ Question on Trigonometry >> Inverse Trigonometry Function

The value of $$\tan \left[ {{{\cos }^{ - 1}}\left( {\frac{4}{5}} \right) + {{\tan }^{ - 1}}\left( {\frac{2}{3}} \right)} \right]$$ is

If we consider only the principle values of the inverse trigonometric functions then the value of $$\tan \left( {{{\cos }^{ - 1}}\frac{1}{{5\sqrt 2 }} - {{\sin }^{ - 1}}\frac{4}{{\sqrt {17} }}} \right)$$ is

The number of real solutions of $${\tan ^{ - 1}}\sqrt {x\left( {x + 1} \right)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = \frac{\pi }{2}$$ is

If $${\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - .....} \right) + {\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - .....} \right) = \frac{\pi }{2}$$ for $$0 < \left| x \right| < \sqrt 2 ,$$ then $$x$$ equals

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Releted MCQ Question on
Trigonometry >> Inverse Trigonometry Function

Practice More Releted MCQ Question on
Inverse Trigonometry Function