Question

If $$\left[ {{{\sin }^{ - 1}}{{\cos }^{ - 1}}{{\sin }^{ - 1}}{{\tan }^{ - 1}}x} \right] = 1,$$       where [.] denotes the greatest integer function, then $$x$$ belongs to the interval

A. $$\left[ {\tan \sin \cos 1,\tan \sin \cos \sin 1} \right]$$  
B. $$\left( {\tan \sin \cos 1,\tan \sin \cos \sin 1} \right)$$
C. $$\left[ {- 1, 1} \right]$$
D. $$\left[ {\sin \cos \tan 1,\sin \cos \sin \tan 1} \right]$$
Answer :   $$\left[ {\tan \sin \cos 1,\tan \sin \cos \sin 1} \right]$$
Solution :
$$\eqalign{ & {\text{We have}},1 \leqslant {\sin ^{ - 1}}{\cos ^{ - 1}}{\sin ^{ - 1}}{\tan ^{ - 1}}x \leqslant \frac{\pi }{2} \cr & \Rightarrow \sin 1 \leqslant {\cos ^{ - 1}}{\sin ^{ - 1}}{\tan ^{ - 1}}x \leqslant 1 \cr & \Rightarrow \cos \sin 1 \geqslant {\sin ^{ - 1}}{\tan ^{ - 1}}x \geqslant \cos 1 \cr & \Rightarrow \sin \cos \sin 1 \geqslant {\tan ^{ - 1}}x \geqslant \sin \cos 1 \cr & \Rightarrow \tan \sin \cos \sin 1 \geqslant x \geqslant \tan \sin \cos 1 \cr & \therefore x \in \left[ {\tan \sin \cos 1,\tan \sin \cos \sin 1} \right] \cr} $$

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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