Question

(1 + cotA + tanA) (sinA - cosA) = (secA / cosec 2 A) - (cosecA / sec 2 A)

Answer 1

Hope it helps my frnds...

Answer 2

Answer:

$(1+cotA+tanA)(sinA-cosA)=\frac{secA}{cosec^{2}A}-\frac{cosecA}{sec^{2}A}$

Step-by-step explanation:

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

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

$=(1+\frac{cosA}{sinA}+\frac{sinA}{cosA})sinA+</p><p>(1+\frac{cosA}{sinA}+\frac{sinA}{cosA})(-cosA)$

$sinA+cosA+\frac{sin^{2}A}{cosA}-cosA-\frac{cos^{2}A}{sinA}-sinA$

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

$=\frac{sin^{2}A}{1}\times \frac{1}{cosA}-\frac{cos^{2}A}{1}\times \frac{1}{sin A}$

$=\frac{secA}{cosec^{2}A}-\frac{cosecA}{sec^{2}A}$

$=RHS$

Therefore,

$(1+cotA+tanA)(sinA-cosA)=\frac{secA}{cosec^{2}A}-\frac{cosecA}{sec^{2}A}$

•••♪