Question

If $\tan \beta = \cos \theta \tan \alpha$ , then ${\tan ^2}\left( {\dfrac{\theta }{2}} \right) =$
\eqalign{ & 1)\dfrac{{\sin \left( {\alpha + \beta } \right)}}{{\sin \left( {\alpha - \beta } \right)}} \cr & 2)\dfrac{{\cos \left( {\alpha - \beta } \right)}}{{\cos \left( {\alpha + \beta } \right)}} \cr & 3)\dfrac{{\sin \left( {\alpha - \beta } \right)}}{{\sin \left( {\alpha + \beta } \right)}} \cr & 4)\dfrac{{\cos \left( {\alpha + \beta } \right)}}{{\cos \left( {\alpha - \beta } \right)}} \cr}

Answer 1

Hint : In the given question, there are only two functions involved, that is $\cos$ and $\tan$ , but with different angles $a$ , $b$ and $\theta$ . But the given options are in $\sin$ and $\cos$ functions. Therefore, we need to express $\tan$ in terms of $\sin$ and $\cos$ functions. Later we can simplify it to bring it down to the form that are given in the options.
The formulas used to solve the given problem are:
\eqalign{ & 1 + \cos 2A = 2{\cos ^2}A \cr & 1 - \cos 2A = 2{\sin ^2}A \cr}

Complete step-by-step answer :
It is given that,
$\tan \beta = \cos \theta \tan \alpha$
We need to find the value of ${\tan ^2}\dfrac{\theta }{2}$
This can be expressed as,
${\tan ^2}\left( {\dfrac{\theta }{2}} \right) = \dfrac{{{{\sin }^2}\left( {\dfrac{\theta }{2}} \right)}}{{{{\cos }^2}\left( {\dfrac{\theta }{2}} \right)}}$
Multiplying and dividing by $2$ , we get
${\tan ^2}\left( {\dfrac{\theta }{2}} \right) = \dfrac{{2{{\sin }^2}\left( {\dfrac{\theta }{2}} \right)}}{{2{{\cos }^2}\left( {\dfrac{\theta }{2}} \right)}}$
By using the above-mentioned formulas, we can write the RHS in the form of $\cos$ function only,
${\tan ^2}\left( {\dfrac{\theta }{2}} \right) = \dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}$
From given, we have,
$\cos \theta = \dfrac{{\tan \beta }}{{\tan \alpha }}$
Substituting this in the above equation, we get,
${\tan ^2}\left( {\dfrac{\theta }{2}} \right) = \left[ {\dfrac{{1 - \left( {\dfrac{{\tan \beta }}{{\tan \alpha }}} \right)}}{{1 + \left( {\dfrac{{\tan \beta }}{{\tan \alpha }}} \right)}}} \right]$
While taking LCM and simplifying, $\tan \alpha$ gets cancelled and we will be left with
${\tan ^2}\left( {\dfrac{\theta }{2}} \right) = \dfrac{{\left[ {\left( {\dfrac{{\sin \alpha }}{{\cos \alpha }} - \dfrac{{\sin \beta }}{{\cos \beta }}} \right)} \right]}}{{\left[ {\left( {\dfrac{{\sin \alpha }}{{\cos \alpha }}} \right) + \left( {\dfrac{{\sin \beta }}{{\cos \beta }}} \right)} \right]}}$
Now, again the terms in the denominator cancel out,
${\tan ^2}\left( {\dfrac{\theta }{2}} \right) = \dfrac{{\left( {\sin \alpha \cos \beta - \sin \beta \cos \alpha } \right)}}{{\left( {\sin \alpha \cos \beta + \sin \beta \cos \alpha } \right)}}$
This can be written as,
${\tan ^2}\left( {\dfrac{\theta }{2}} \right) = \dfrac{{\sin \left( {\alpha - \beta } \right)}}{{\sin \left( {\alpha + \beta } \right)}}$
Therefore, the final answer is, $\dfrac{{\sin \left( {\alpha - \beta } \right)}}{{\sin \left( {\alpha + \beta } \right)}}$
Hence, option (3) is the correct answer.
So, the correct answer is “Option 3”.

Note : When we look at a function, we need to decide which formula will be suitable to apply for it. Then we can add, subtract, multiply or divide the necessary terms. Learn the basic formulas and apply them as required for each step. Make sure that you simplify the answer according to the options. We can do so, by choosing the appropriate formulas as we move on with the steps. The options look similar and confusing, be very careful while choosing the right one.