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 .