Question

If $2\sec 2\alpha = \tan \beta + \cot \beta $,then what is the one possible value of $\alpha + \beta $?
A. $\dfrac{\pi }{4}$
B. $\dfrac{\pi }{2}$
C. $\pi $
D. $n\pi - \dfrac{\pi }{4},n \in I$

Answer 1

Hint: We will use basic trigonometric formulas to simplify RHS and LHS of a given equation and finally we will solve the equation with simplified parts.

Formula Used:
$\sin 2A = 2\sin A\cos A$
${\sec ^2}A = 1 + {\tan ^2}A$
$\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta $

Complete step by step solution:
Given -$2\sec 2\alpha = \tan \beta + \cot \beta $
RHS =$\tan \beta + \cot \beta $
$ \Rightarrow $ RHS =$\tan \beta + \dfrac{1}{{\tan \beta }}$
(As $\cot \beta = \dfrac{1}{{\tan \beta }}$)
Taking LCM and simplifying the expression
RHS=$\dfrac{{\tan \beta *\tan \beta + 1}}{{\tan \beta }}$
RHS =$\dfrac{{1 + {{\tan }^2}\beta }}{{\tan \beta }}$
But we know ${\sec ^2}A = 1 + {\tan ^2}A$
So, RHS=$\dfrac{{{{\sec }^2}\beta }}{{\tan \beta }}$
But $\sec A = \dfrac{1}{{\cos A}}$
Hence, RHS=$\dfrac{1}{{{{\cos }^2}\beta \tan \beta }}$
$ \Rightarrow $ RHS =$\dfrac{1}{{{{\cos }^2}\beta \dfrac{{\sin \beta }}{{\cos \beta }}}}$
$ \Rightarrow $ RHS=$\dfrac{1}{{\cos \beta \sin \beta }}$ (1)
Now taking LHS,
LHS=$2\sec 2\alpha $
Using formula$\sec A = \dfrac{1}{{\cos A}}$
$ \Rightarrow $ LHS=$\dfrac{2}{{\cos 2\alpha }}$(2)
Now comparing equation (1) and (2)
$\dfrac{2}{{\cos 2\alpha }} = \dfrac{1}{{\cos \beta \sin \beta }}$
 On cross multiplying terms
$2\sin \beta \cos \beta = \cos 2\alpha $
Using the formula $\sin 2A = 2\sin A\cos A$
$\sin 2\beta = \cos 2\alpha $
We know that, $\sin (90 - \theta ) = \cos \theta $
Hence $\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta $
So $\cos \left( {\dfrac{\pi }{2} - 2\beta } \right)$=$\cos 2\alpha $
Thus $2\alpha = \dfrac{\pi }{2} - 2\beta $
$\Rightarrow 2\alpha + 2\beta = \dfrac{\pi }{2}$
$\alpha + \beta = \dfrac{\pi }{4}$

Option ‘A’ is correct

Note: We often make mistakes while solving trigonometric equations. For example, the equation given here is $\cos \left( {\dfrac{\pi }{2} - 2\beta } \right)=\cos 2\alpha $. Writing $2\alpha = \dfrac{\pi }{2} - 2\beta$ is an incomplete solution since this is a trigonometric equation and there can be various values of $\beta$ for which following equation can be satisfied due to the periodic properties of function.
We know that the general solution for equation $\cos A = \cos B$ is given by $A = 2n\pi \pm B$.
Now our equation is $\cos \left( {\dfrac{\pi }{2} - 2\beta } \right)$=$\cos 2\alpha $
Taking $A = \dfrac{\pi }{2} - 2\beta $and $B = 2\alpha $,the general solution can be written as
$\dfrac{\pi }{2} - 2\beta = 2n\pi \pm 2\alpha $
Taking positive sign to simplify the equation
$\dfrac{\pi }{2} - 2\beta = 2n\pi + 2\alpha $
$ \Rightarrow \dfrac{\pi }{2} - 2n\pi = 2\alpha + 2\beta $
$ \Rightarrow 2\alpha + 2\beta = \dfrac{\pi }{2} - 2n\pi $
$ \Rightarrow \alpha + \beta = \dfrac{\pi }{4} - n\pi $
This is the general solution, when we put $n = 0$, we will get $\alpha + \beta = \dfrac{\pi }{4}$,which is one of the possible solutions.