JEE MAIN - Mathematics (2019 - 8th April Morning Slot - No. 1)
If $$\alpha = {\cos ^{ - 1}}\left( {{3 \over 5}} \right)$$, $$\beta = {\tan ^{ - 1}}\left( {{1 \over 3}} \right)$$ where $$0 < \alpha ,\beta < {\pi \over 2}$$ , then $$\alpha $$ - $$\beta $$ is equal to :
$${\tan ^{ - 1}}\left( {{9 \over {14 }}} \right)$$
$${\sin ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)$$
$${\cos ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)$$
$${\tan ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)$$
Explanation
Here $$\cos \alpha = {3 \over 5}$$
$$ \therefore $$ $$\tan \alpha = {4 \over 3}$$
and $$\tan \beta = {1 \over 3}$$
We know,
$$\tan \left( {\alpha - \beta } \right) = {{\tan \alpha - \tan \beta } \over {1 + \tan \alpha .\tan \beta }}$$
= $${{{4 \over 3} - {1 \over 3}} \over {1 + {4 \over 3}.{1 \over 3}}}$$ = $${9 \over {13}}$$
$$ \therefore $$ $$\left( {\alpha - \beta } \right)$$ = $${\tan ^{ - 1}}\left( {{9 \over {13}}} \right)$$ = $${\sin ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)$$
$$ \therefore $$ $$\tan \alpha = {4 \over 3}$$
and $$\tan \beta = {1 \over 3}$$
We know,
$$\tan \left( {\alpha - \beta } \right) = {{\tan \alpha - \tan \beta } \over {1 + \tan \alpha .\tan \beta }}$$
= $${{{4 \over 3} - {1 \over 3}} \over {1 + {4 \over 3}.{1 \over 3}}}$$ = $${9 \over {13}}$$
$$ \therefore $$ $$\left( {\alpha - \beta } \right)$$ = $${\tan ^{ - 1}}\left( {{9 \over {13}}} \right)$$ = $${\sin ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)$$
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