Question

If $\sin 2a=\lambda \sin 2b$, prove that $\dfrac{\tan (a+b)}{\tan (a-b)}=\dfrac{\lambda +1}{\lambda -1}$

Answer 1

We have an equation given as $\sin 2a=\lambda \sin 2b$ so we first calculate value of $\dfrac{\sin 2a}{\sin 2b}=\lambda$ then putting its value in formula $\dfrac{\lambda +1}{\lambda -1}$ as this is the expression in right hand side
We know some trigonometric formulas like $\sin a+\sin b=2\sin \dfrac{a+b}{2}\cos \dfrac{a-b}{2}$
And $\sin a-\sin b=2\sin \dfrac{a-b}{2}\cos \dfrac{a+b}{2}$ and $\dfrac{\sin (a-b)}{\cos (a-b)}=\tan (a-b)$ , $\dfrac{\sin (a+b)}{\cos (a+b)}=\tan (a+b)$ we will apply it in our question and find the desired result.

Complete step-by-step answer:
We are given an equation $\sin 2a=\lambda \sin 2b$
Which can be written as $\dfrac{\sin 2a}{\sin 2b}=\lambda$
On RHS side of question we are given an expression $\dfrac{\lambda +1}{\lambda -1}$ for that we can calculate the value of $\dfrac{\lambda +1}{\lambda -1}$ by putting value of $\lambda$ in it
On putting the value, we get $\dfrac{\lambda +1}{\lambda -1}=\dfrac{\dfrac{\sin 2a}{\sin 2b}+1}{\dfrac{\sin 2a}{\sin 2b}-1}$, on solving further we can write it as
$\dfrac{\lambda +1}{\lambda -1}=\dfrac{\sin 2a+\sin 2b}{\sin 2a-\sin 2b}$
Now using formula $\sin a+\sin b=2\sin \dfrac{a+b}{2}\cos \dfrac{a-b}{2}$
And $\sin a-\sin b=2\sin \dfrac{a-b}{2}\cos \dfrac{a+b}{2}$, In our equation we substitute a and b with 2a and 2b our equation will transform to
$\dfrac{\sin 2a+\sin 2b}{\sin 2a-\sin 2b}=\dfrac{2\sin \dfrac{2a+2b}{2}\cos \dfrac{2a-2b}{2}}{2\sin \dfrac{2a-2b}{2}\cos \dfrac{2a+2b}{2}}$
On solving it further our expression looks like
$\dfrac{2\sin \dfrac{2a+2b}{2}\cos \dfrac{2a-2b}{2}}{2\sin \dfrac{2a-2b}{2}\cos \dfrac{2a+2b}{2}}=\dfrac{2\sin (a+b)\cos (a-b)}{2\sin (a-b)\cos (a+b)}$
On further solving we comes to our final answer as
We know one basic formula of trigonometric which is $\dfrac{\sin (a-b)}{\cos (a-b)}=\tan (a-b)$ and $\dfrac{\sin (a+b)}{\cos (a+b)}=\tan (a+b)$ applying this equation in our expression it will look like
$\dfrac{2\sin (a+b)\cos (a-b)}{2\sin (a-b)\cos (a+b)}$
Which on further solving equals to $=\dfrac{\tan (a+b)}{\tan (a-b)}$
$\dfrac{\tan (a+b)}{\tan (a-b)}$ is our left-hand side in the given question?
Hence, we proved, LHS=RHS

Note: We can also try this question from different approach like solving the LHS part first for example equation this $\dfrac{\tan (a+b)}{\tan (a-b)}=\dfrac{\lambda +1}{\lambda -1}$ can be written as $\dfrac{\tan (a+b)}{\tan (a-b)}=\dfrac{\dfrac{\sin (a+b)}{\cos (a+b)}}{\dfrac{\sin (a-b)}{\cos (a-b)}}$
On solving further, it can be written as $\dfrac{\tan (a+b)}{\tan (a-b)}=\dfrac{\sin (a+b)cos(a-b)}{\sin (a-b)cos(a+b)}$ and now expanding using trigonometric formulas but as you can see that this method is quite tedious.
Most of the students don’t remember trigonometric formulas and make mistakes while writing it, so they learn all the formulas very properly.