Question

If $\cos 2\alpha = \dfrac{{\left( {3\cos 2\beta - 1} \right)}}{{3 - \cos 2\beta }}$ , then $\tan \alpha$ is equal to
A.$\sqrt 2 \tan \beta$
B.$\tan \beta$
C.$\sin 2\beta$
D.$\sqrt 2 \cot \beta$

Answer 1

Hint : : In the given question we have to use the different identities of trigonometric ratios of compound angles we will use the formula $\cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}$ in numerator and denominator of the given question. We have used this formula as we must find the value of $\tan \alpha$ and then compare it again with the identity of $\cos 2x$ .

Complete step-by-step answer :
Given : $\cos 2\alpha = \dfrac{{\left( {3\cos 2\beta - 1} \right)}}{{3 - \cos 2\beta }}$ …..equation (a) .
Now using the identity $\cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}$ in equation (a) we get ,
$\cos 2\alpha = \dfrac{{\left( {3\left( {\dfrac{{1 - {{\tan }^2}\beta }}{{1 + {{\tan }^2}\beta }}} \right) - 1} \right)}}{{3 - \left( {\dfrac{{1 - {{\tan }^2}\beta }}{{1 + {{\tan }^2}\beta }}} \right)}}$
On solving we get ,
$\cos 2\alpha = \dfrac{{\left( {\dfrac{{3 - 3{{\tan }^2}\beta }}{{1 + {{\tan }^2}\beta }}} \right) - 1}}{{3 - \left( {\dfrac{{1 - {{\tan }^2}\beta }}{{1 + {{\tan }^2}\beta }}} \right)}}$
$\cos 2\alpha = \dfrac{{\left( {\dfrac{{3 - 3{{\tan }^2}\beta - 1 - {{\tan }^2}\beta }}{{1 + {{\tan }^2}\beta }}} \right)}}{{\left( {\dfrac{{3 + 3{{\tan }^2}\beta - 1 + {{\tan }^2}\beta }}{{1 + {{\tan }^2}\beta }}} \right)}}$
On solving further we get ,
$\cos 2\alpha = \dfrac{{\left( {\dfrac{{2 - 4{{\tan }^2}\beta }}{{1 + {{\tan }^2}\beta }}} \right)}}{{\left( {\dfrac{{2 + 4{{\tan }^2}\beta }}{{1 + {{\tan }^2}\beta }}} \right)}}$
Now cancelling out the denominator we get ,
$\cos 2\alpha = \dfrac{{2 - 4{{\tan }^2}\beta }}{{2 + 4{{\tan }^2}\beta }}$
Taking $2$ common and cancelling it out we get ,
$\cos 2\alpha = \dfrac{{1 - 2{{\tan }^2}\beta }}{{1 + 2{{\tan }^2}\beta }}$ … equation ( b ) .
Now , we know that $\cos 2\alpha = \dfrac{{1 - {{\tan }^2}\alpha }}{{1 + {{\tan }^2}\alpha }}$ , on comparing with equation (b) we get ,
${\tan ^2}\alpha = 2{\tan ^2}\beta$
Taking square root on both sides we get ,
$\tan \alpha = \sqrt 2 \tan \beta$ .
Therefore , option ( 1 ) is the correct answer for this question .
So, the correct answer is “Option A”.

Note : Trigonometry is all about angles and their measurement . When discussing the various trigonometric functions, we keep in mind the formula of compound angles to give accurate results. A compound cut comprises two angles . The Trigonometric identities should be remembered to solve questions related to identities . Moreover , questions related to identities can be solved using different ways . Also the solution of the question can be different , it depends on the identity you are using in your question . . You also have to remember the values of trigonometric ratios at different angles . These questions are short and tricky .