JEE MAIN - Mathematics (2020 - 8th January Evening Slot - No. 6)
If $${{\sqrt 2 \sin \alpha } \over {\sqrt {1 + \cos 2\alpha } }} = {1 \over 7}$$ and $$\sqrt {{{1 - \cos 2\beta } \over 2}} = {1 \over {\sqrt {10} }}$$
$$\alpha ,\beta \in \left( {0,{\pi \over 2}} \right)$$ then tan($$\alpha $$ + 2$$\beta $$) is equal to _____.
$$\alpha ,\beta \in \left( {0,{\pi \over 2}} \right)$$ then tan($$\alpha $$ + 2$$\beta $$) is equal to _____.
Answer
1
Explanation
$${{\sqrt 2 \sin \alpha } \over {\sqrt {1 + \cos 2\alpha } }} = {1 \over 7}$$
$$ \Rightarrow $$ $${{\sqrt 2 \sin \alpha } \over {\sqrt {2{{\cos }^2}\alpha } }}$$ = $${1 \over 7}$$
$$ \Rightarrow $$ $${{\sqrt 2 \sin \alpha } \over {\sqrt 2 \cos \alpha }}$$ = $${1 \over 7}$$
$$ \Rightarrow $$ tan$$\alpha $$ = $${1 \over 7}$$
Also given $$\sqrt {{{1 - \cos 2\beta } \over 2}} = {1 \over {\sqrt {10} }}$$
$$ \Rightarrow $$ $${{\sqrt 2 \sin \beta } \over {\sqrt 2 }}$$ = $${1 \over {\sqrt {10} }}$$
$$ \Rightarrow $$ sin $$\beta $$ = $${1 \over {\sqrt {10} }}$$
$$ \therefore $$ tan $$\beta $$ = $${1 \over 3}$$
$$\tan 2\beta = {{2\tan \beta } \over {1 - {{\tan }^2}\beta }}$$
= $${{2\left( {{1 \over 3}} \right)} \over {1 - {1 \over 9}}}$$ = $${3 \over 4}$$
$$ \therefore $$ tan($$\alpha $$ + 2$$\beta $$) = $${{\tan \alpha + \tan 2\beta } \over {1 - \tan \alpha .\tan 2\beta }}$$
= $${{{1 \over 7} + {3 \over 4}} \over {1 - {1 \over 7}.{3 \over 4}}}$$
= $${{25} \over {25}}$$ = 1
$$ \Rightarrow $$ $${{\sqrt 2 \sin \alpha } \over {\sqrt {2{{\cos }^2}\alpha } }}$$ = $${1 \over 7}$$
$$ \Rightarrow $$ $${{\sqrt 2 \sin \alpha } \over {\sqrt 2 \cos \alpha }}$$ = $${1 \over 7}$$
$$ \Rightarrow $$ tan$$\alpha $$ = $${1 \over 7}$$
Also given $$\sqrt {{{1 - \cos 2\beta } \over 2}} = {1 \over {\sqrt {10} }}$$
$$ \Rightarrow $$ $${{\sqrt 2 \sin \beta } \over {\sqrt 2 }}$$ = $${1 \over {\sqrt {10} }}$$
$$ \Rightarrow $$ sin $$\beta $$ = $${1 \over {\sqrt {10} }}$$
$$ \therefore $$ tan $$\beta $$ = $${1 \over 3}$$
$$\tan 2\beta = {{2\tan \beta } \over {1 - {{\tan }^2}\beta }}$$
= $${{2\left( {{1 \over 3}} \right)} \over {1 - {1 \over 9}}}$$ = $${3 \over 4}$$
$$ \therefore $$ tan($$\alpha $$ + 2$$\beta $$) = $${{\tan \alpha + \tan 2\beta } \over {1 - \tan \alpha .\tan 2\beta }}$$
= $${{{1 \over 7} + {3 \over 4}} \over {1 - {1 \over 7}.{3 \over 4}}}$$
= $${{25} \over {25}}$$ = 1
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