Question

Prove that:

$\tan (x-y)=\left[\frac{\tan \left(x\right)-\tan \left(y\right)}{1+\tan \left(x\right)\tan \left(y\right)}\right]$

Answer 1

STEP 1 : Solving the Left Hand Side (LHS) of the equation

Taking the LHS and solving we get,

$\Rightarrow \tan (x-y)$

$=\frac{\sin (x-y)}{\cos (x-y)}$ $...\left(1\right)$

We know that,

$\sin (x-y)=\sin \left(x\right)\cos \left(y\right)-\cos \left(x\right)\sin \left(y\right)$ and

$\cos (x-y)=\cos \left(x\right)\cos \left(y\right)+\sin \left(x\right)\sin \left(y\right)$

Substituting the values of $\sin (x-y)$ and $\cos (x-y)$ in equation $\left(1\right)$ we get,

$=\frac{\sin \left(x\right)\cos \left(y\right)-\cos \left(x\right)\sin \left(y\right)}{\cos \left(x\right)\cos \left(y\right)+\sin \left(x\right)\sin \left(y\right)}$

Divide numerator and denominator by $\cos \left(x\right)\cos \left(y\right)$

$=\frac{{\displaystyle \frac{\sin \left(x\right)\cos \left(y\right)-\cos \left(x\right)\sin \left(y\right)}{\cos \left(x\right)\cos \left(y\right)}}}{{\displaystyle \frac{\cos \left(x\right)\cos \left(y\right)+\sin \left(x\right)\sin \left(y\right)}{\cos \left(x\right)\cos \left(y\right)}}}$

$=\frac{{\displaystyle \frac{\sin \left(x\right)\cos \left(y\right)}{\cos \left(x\right)\cos \left(y\right)}}-{\displaystyle \frac{\cos \left(x\right)\sin \left(y\right)}{\cos \left(x\right)\cos \left(y\right)}}}{{\displaystyle \frac{\cos \left(x\right)\cos \left(y\right)}{\cos \left(x\right)\cos \left(y\right)}}+{\displaystyle \frac{\sin \left(x\right)\sin \left(y\right)}{\cos \left(x\right)\cos \left(y\right)}}}$

$=\frac{\tan \left(x\right)-{\displaystyle \tan \left(y\right)}}{{\displaystyle 1}+{\displaystyle \tan \left(x\right)\tan \left(y\right)}}=RHS$

i.e.$\tan (x-y)=\left[\frac{\tan \left(x\right)-\tan \left(y\right)}{1+\tan \left(x\right)\tan \left(y\right)}\right]$

Hence proved.