JEE MAIN - Mathematics (2020 - 3rd September Morning Slot - No. 2)
2$$\pi $$ - $$\left( {{{\sin }^{ - 1}}{4 \over 5} + {{\sin }^{ - 1}}{5 \over {13}} + {{\sin }^{ - 1}}{{16} \over {65}}} \right)$$ is equal to :
$${{7\pi } \over 4}$$
$${{5\pi } \over 4}$$
$${{3\pi } \over 2}$$
$${\pi \over 2}$$
Explanation
$$2\pi - \left( {{{\sin }^{ - 1}}{4 \over 5} + {{\sin }^{ - 1}}{5 \over {13}} + {{\sin }^{ - 1}}{{16} \over {65}}} \right)$$
$$ = 2\pi - \left( {{{\tan }^{ - 1}}{4 \over 3} + {{\tan }^{ - 1}}{5 \over {12}} + {{\tan }^{ - 1}}{{16} \over {63}}} \right)$$
$$ = 2\pi - \left\{ {{{\tan }^{ - 1}}\left( {{{{4 \over 3} + {5 \over {12}}} \over {1 - {4 \over 3}.{5 \over {12}}}}} \right) + {{\tan }^{ - 1}}{{16} \over {63}}} \right\}$$
$$ = 2\pi - \left( {{{\tan }^{ - 1}}{{63} \over {16}} + {{\tan }^{ - 1}}{{16} \over {63}}} \right)$$
= $$2\pi - \left( {{{\tan }^{ - 1}}{{63} \over {16}} + {{\cot }^{ - 1}}{{63} \over {16}}} \right)$$
$$ = 2\pi - {\pi \over 2}$$
$$ = {{3\pi } \over 2}$$
$$ = 2\pi - \left( {{{\tan }^{ - 1}}{4 \over 3} + {{\tan }^{ - 1}}{5 \over {12}} + {{\tan }^{ - 1}}{{16} \over {63}}} \right)$$
$$ = 2\pi - \left\{ {{{\tan }^{ - 1}}\left( {{{{4 \over 3} + {5 \over {12}}} \over {1 - {4 \over 3}.{5 \over {12}}}}} \right) + {{\tan }^{ - 1}}{{16} \over {63}}} \right\}$$
$$ = 2\pi - \left( {{{\tan }^{ - 1}}{{63} \over {16}} + {{\tan }^{ - 1}}{{16} \over {63}}} \right)$$
= $$2\pi - \left( {{{\tan }^{ - 1}}{{63} \over {16}} + {{\cot }^{ - 1}}{{63} \over {16}}} \right)$$
$$ = 2\pi - {\pi \over 2}$$
$$ = {{3\pi } \over 2}$$
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