Question

Prove that $\dfrac{\sin (x+y)}{\sin (x-y)}=\dfrac{\tan x+\tan y}{\tan x-\tan y}$ .

Answer 1

Hint: Try to simplify the left-hand side of the equation given in the question. Start by using the formula of sin(A+B) and sin(A-B). Finally, divide the numerator and the denominator by cosxcosy to solve the left-hand side of the equation given in the question.

Complete step-by-step answer:
Let us start the simplification of the left-hand side of the equation.
$\dfrac{\sin \left( x+y \right)}{\sin \left( x-y \right)}$
Now we know that $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ and $\text{sin}\left( A+B \right)=\sin A\cos B+\cos A\sin B$ . On using this in our expression, we get
$\dfrac{\sin x\cos y+\cos x\sin y}{\sin x\cos y-\cos x\sin y}$
Now we will divide both the numerator and the denominator of the expression by cosxcosy. On doing so, we get
$\dfrac{\dfrac{\sin x\cos y}{\cos x\cos y}+\dfrac{\cos x\sin y}{\cos x\cos y}}{\dfrac{\sin x\cos y}{\cos x\cos y}+\dfrac{\cos x\sin y}{\cos x\cos y}}$
We know that $\tan A=\dfrac{\sin A}{\cos A}$ .
$\therefore \dfrac{\dfrac{\sin x}{\cos x}+\dfrac{\sin y}{\cos y}}{\dfrac{\sin x}{\cos x}-\dfrac{\sin y}{\cos y}}$
$=\dfrac{\tan x+\tan y}{\tan x-\tan y}$
The left-hand side of the equation given in the question is equal to the right-hand side of the equation. Hence, we can say that we have proved the equation given in the question.

Note: Be careful about the calculation and the signs while opening the brackets. The general mistake that a student can make is 1+x-(x-1)=1+x-x-1. Also, be careful about the signs in the formula of sin(A-B) and sin(A+B). Whenever you are dealing with an expression having cotangents, secant, and cosecant involved, it is better to convert it to an equivalent expression in terms of sine, cosine, and tangent, as most of the formulas we know are valid for sine, cosine, and tangents only.