Question

The value of $${\cot ^{ - 1}}7 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18$$      is

A. $$\pi$$
B. $$\frac{\pi }{2}$$
C. $${\cot ^{ - 1}}5$$
D. $${\cot ^{ - 1}}3$$  
Answer :   $${\cot ^{ - 1}}3$$
Solution :
$$\eqalign{ & {\text{we have,}}\,\,{\cot ^{ - 1}}7 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18 \cr & {\tan ^{ - 1}}\frac{1}{7} + {\tan ^{ - 1}}\frac{1}{8} + {\tan ^{ - 1}}\frac{1}{{18}} \cr & = {\tan ^{ - 1}}\left( {\frac{{\frac{1}{7} + \frac{1}{8}}}{{1 - \frac{1}{7} \times \frac{1}{8}}}} \right) + {\tan ^{ - 1}}\frac{1}{{18}} \cr & = {\tan ^{ - 1}}\frac{{15}}{{55}}{\tan ^{ - 1}}\frac{1}{{18}}\,\,\left( {\because \frac{1}{7} \cdot \frac{1}{8} < 1} \right) \cr & {\tan ^{ - 1}}\frac{3}{{11}} + {\tan ^{ - 1}}\frac{1}{{18}} \cr & = {\tan ^{ - 1}}\left( {\frac{{\frac{3}{{11}} + \frac{1}{{18}}}}{{1 - \frac{3}{{11}} \times \frac{1}{{18}}}}} \right)\left( {\because \frac{3}{{11}} \cdot \frac{1}{{18}} < 1} \right) \cr & = {\tan ^{ - 1}}\frac{{65}}{{195}} = {\tan ^{ - 1}}\frac{1}{3} = {\cot ^{ - 1}}3 \cr} $$

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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