Question

Total number of positive integral value $$'n'$$ so that the equations $${\cos ^{ - 1}}x + {\left( {{{\sin }^{ - 1}}y} \right)^2} = \frac{{n{\pi ^2}}}{4}\,$$     and $${\left( {{{\sin }^{ - 1}}y} \right)^2} - {\cos ^{ - 1}}x = \frac{{{\pi ^2}}}{{16}}\,$$     are consistent, is equal to

A. 1  
B. 4
C. 3
D. 2
Answer :   1
Solution :
$$\eqalign{ & {\text{We have, }}2{\left( {{{\sin }^{ - 1}}y} \right)^2} = \frac{{4n + 1}}{{16}}{\pi ^2} \cr & \Rightarrow 0 \leqslant \frac{{4n + 1}}{{32}}{\pi ^2} \leqslant \frac{{{\pi ^2}}}{4} \cr & {\text{Also, }}2\left( {{{\cos }^{ - 1}}x} \right) = \frac{{4n - 1}}{{16}}{\pi ^2} \cr & \Rightarrow - \frac{1}{4} \leqslant n \leqslant \frac{7}{4} \cr & {\text{Also, }}2\left( {{{\cos }^{ - 1}}x} \right) = \frac{{4n - 1}}{{16}}{\pi ^2} \cr & \Rightarrow 0 \leqslant \frac{{4n - 1}}{{32}}{\pi ^2} \leqslant \pi \cr & \Rightarrow \frac{1}{4} \leqslant n \leqslant \frac{8}{\pi } + \frac{1}{4} \cr & \Rightarrow n = 1 \cr} $$

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
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. $$\frac{{3}}{29}$$
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
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. $$- 1$$

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Inverse Trigonometry Function


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