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2$$\pi $$ - $$\left( {{{\sin }^{ - 1}}{4 \over 5} + {{\sin }^{ - 1}}{5 \over {13}} + {{\sin }^{ - 1}}{{16} \over {65}}} \right)$$ is equal to :

JEE · Math · previous-year question

  1. A.$${{7\pi } \over 4}$$
  2. B.$${{5\pi } \over 4}$$
  3. C.$${{3\pi } \over 2}$$correct
  4. D.$${\pi \over 2}$$

Answer

C. $${{3\pi } \over 2}$$

Explanation

$$2\pi - \left( {{{\sin }^{ - 1}}{4 \over 5} + {{\sin }^{ - 1}}{5 \over {13}} + {{\sin }^{ - 1}}{{16} \over {65}}} \right)$$ $$ = 2\pi - \left( {{{\tan }^{ - 1}}{4 \over 3} + {{\tan }^{ - 1}}{5 \over {12}} + {{\tan }^{ - 1}}{{16} \over {63}}} \right)$$ $$ = 2\pi - \left\{ {{{\tan }^{ - 1}}\left( {{{{4 \over 3} + {5 \over {12}}} \over {1 - {4 \over 3}.{5 \over {12}}}}} \right) + {{\tan }^{ - 1}}{{16} \over {63}}} \right\}$$ $$ = 2\pi - \left( {{{\tan }^{ - 1}}{{63} \over {16}} + {{\tan }^{ - 1}}{{16} \over {63}}} \right)$$ = $$2\pi - \left( {{{\tan }^{ - 1}}{{63} \over {16}} + {{\cot }^{ - 1}}{{63} \over {16}}} \right)$$ $$ = 2\pi - {\pi \over 2}$$ $$ = {{3\pi } \over 2}$$

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