JEE MAIN - Mathematics (2010 - No. 21)
Let $$\cos \left( {\alpha + \beta } \right) = {4 \over 5}$$ and $$\sin \,\,\,\left( {\alpha - \beta } \right) = {5 \over {13}},$$ where $$0 \le \alpha ,\,\beta \le {\pi \over 4}.$$
Then $$tan\,2\alpha $$ =
Then $$tan\,2\alpha $$ =
$${56 \over 33}$$
$${19 \over 12}$$
$${20 \over 7}$$
$${25 \over 16}$$
Explanation
$$\cos \left( {\alpha + \beta } \right) = {4 \over 5} \Rightarrow \tan \left( {\alpha + \beta } \right) = {3 \over 4}$$
$$\sin \left( {\alpha - \beta } \right) = {5 \over {13}} \Rightarrow \tan \left( {\alpha - \beta } \right) = {5 \over {12}}$$
$$\tan 2\alpha = \tan \left[ {\left( {\alpha + \beta } \right) + \left( {\alpha - \beta } \right)} \right]$$
$$ = {{{3 \over 4} + {5 \over {12}}} \over {1 - {3 \over 4}.{5 \over {12}}}} = {{56} \over {33}}$$
$$\sin \left( {\alpha - \beta } \right) = {5 \over {13}} \Rightarrow \tan \left( {\alpha - \beta } \right) = {5 \over {12}}$$
$$\tan 2\alpha = \tan \left[ {\left( {\alpha + \beta } \right) + \left( {\alpha - \beta } \right)} \right]$$
$$ = {{{3 \over 4} + {5 \over {12}}} \over {1 - {3 \over 4}.{5 \over {12}}}} = {{56} \over {33}}$$
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