Question

The solution set of the equation $${\cos ^{ - 1}}x - {\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {1 - x} \right)$$      is

A. $$\left[ { - 1,1} \right]$$
B. $$\left[ {0,\frac{1}{2}} \right]$$
C. $$\left[ { - 1,0} \right]$$
D. None of these  
Answer :   None of these
Solution :
$${\sin ^{ - 1}}x,{\cos ^{ - 1}}x\,$$   exist for $$ - 1 \leqslant x \leqslant 1.$$   But for $${\sin ^{ - 1}}\left( {1 - x} \right)$$   we must have $$ - 1 \leqslant 1 - x \leqslant 1,\,{\text{i}}{\text{.e}}{\text{., }}0 \leqslant x \leqslant 2.$$      So the equation may hold for $$0 \leqslant x \leqslant 1.$$  Therefore, the options $$A, B$$  and $$C$$ are incorrect.

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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