Question

Prove that :
$(1+\cot A+\tan A)(\sin A-\cos A)=\sin A\tan A-\cot A\cos A$.

Answer 1

Let

$L.H.S=$ $(1+cotA+tanA)(sinA-cosA)$

$=\left(1+\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}\right)\left(\sin A-\cos A\right)$

$=\left(\frac{\sin A\cos A+{\cos }^{2}A+{\sin }^{2}A}{\sin A\cos A}\right)\left(\sin A-\cos A\right)$

$=\left(1+\sin A\cos A\right)\left(\frac{\sin A-\cos A}{\sin A\cos A}\right)$

Now,

$R.H.S=\sin A\tan A-\cot A\cos A$

$=\sin A\frac{\sin A}{\cos A}-\frac{\cos A}{\sin A}\cos A$

$=\frac{{\sin }^{2}A}{\cos A}-\frac{{\cos }^{2}A}{\sin A}$

$=\frac{{\sin }^{3}A-{\cos }^{3}A}{\sin A\cos A}$

$=\frac{\left(\sin A-\cos A\right)}{\sin A\cos A}\left({\sin }^{2}A+{\cos }^{2}A+\sin A\cos A\right)$

$=\left(1+\sin A\cos A\right)\frac{\left(\sin A-\cos A\right)}{\sin A\cos A}$

Hence,

$L.H.S=R.H.S$

proved