Question

Prove the following identity :
$(1+\cot A+\tan A)(\sin A-\cos A)=\frac{\sec A}{{\csc }^{2}A}-\frac{\csc A}{{\sec }^{2}A}$

Answer 1

prove :

$(1+cotA+tanA)(sinA-cosA)=\frac{secA}{cos{e}^{2}A}-\frac{cosecA}{se{c}^{2}A}$

LHS $=(1+cotA+tanA)(sinA-cosA)$

$=(1+\frac{cosA}{sinA}+\frac{sinA}{cosA})(sinA-cosA)$

$=\left(\frac{sinA.cosA+co{s}^{2}A+si{n}^{2}A}{sinA.cosA}\right)(sinA-cosA)$

$=\left(\frac{sinA.cosA+1}{sinA.cosA}\right)(sinA-cosA)$

RHS $=\frac{secA}{cose{c}^{2}A}-\frac{cosecA}{se{c}^{2}A}$

$=\frac{1/cosA}{1/si{n}^{2}A}-\frac{1/sinA}{1/co{s}^{2}A}$

$=\frac{si{n}^{2}A}{cosA}-\frac{co{s}^{2}A}{sinA}$

$=\frac{si{n}^{3}A-co{s}^{3}A}{sinA.cosA}$

$=\frac{(sinA-cosA)(si{n}^{2}A+co{s}^{2}A+sinA.cosA)}{sinA.cosA}$

$=\frac{(sinA-cosA)(1+sinA.cosA)}{sinA.cosA}$

hence, LHS = RHS