If [ = ^ - 1( 3 5 )], [ = ^ - 1( 1 3 )] where [0

Question

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 :

Accepted Answer

Here $$\cos \alpha  = {3 \over 5}$$

$$ \therefore $$ $$\tan \alpha  = {4 \over 3}$$

and $$\tan \beta  = {1 \over 3}$$

We know,

$$\tan \left( {\alpha  - \beta } \right) = {{\tan \alpha  - \tan \beta } \over {1 + \tan \alpha .\tan \beta }}$$

= $${{{4 \over 3} - {1 \over 3}} \over {1 + {4 \over 3}.{1 \over 3}}}$$ = $${9 \over {13}}$$

$$ \therefore $$ $$\left( {\alpha  - \beta } \right)$$ = $${\tan ^{ - 1}}\left( {{9 \over {13}}} \right)$$ = $${\sin ^{ - 1}}\left( {{9 \over {5\sqrt {10} }}} \right)$$

Solutions · 1