the formula cos 1 1 x 21 x 2 2tan 1x holds only for

The formula $${\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}} = 2{\tan ^{ - 1}}x$$ holds only for

A. $$x \in R$$

B. $$\left| x \right| \leqslant 1$$

C. $$x \in \left( { - 1,1} \right]$$

D. $$x \in \left[ {1, + \infty } \right)$$

Answer : $$x \in \left[ {1, + \infty } \right)$$

Solution :
If $$x = - 1,$$ L.H.S $$ = \frac{\pi }{2},$$ R.H.S. $$ = 2 \times \left( { - \frac{\pi }{2}} \right).$$ So, the formula does not hold.
If $$x < - 1,$$ the angle on the L.H.S. is in the second quadrant while the angle on the R.H.S. is 2 $$ \times $$ (angle in the fourth quadrant), which cannot be equal.
If $$x > 1,$$ the angle on the L.H.S. is in the second quadrant while the angle on the R.H.S. is 2 $$ \times $$ (angle in the first quadrant) and these two may be equal.
If $$- 1 < x < 0,$$ the angle on the L.H.S. is positive and that on the R.H.S. is negative and the two cannot be equal.

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

View Answer

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

The formula $${\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}} = 2{\tan ^{ - 1}}x$$ holds only for

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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Inverse Trigonometry Function

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The formula $${\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}} = 2{\tan ^{ - 1}}x$$ holds only for

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

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Releted MCQ Question on
Trigonometry >> Inverse Trigonometry Function

Practice More Releted MCQ Question on
Inverse Trigonometry Function