Question

prove that sin2A÷sina-cos2A÷cosa=secA

Answer 1

GIVEN :

Prove that $\frac{sin2A}{sinA}-\frac{cos2A}{cosA}=secA$

TO FIND :

The given equation is true and check the equality.

SOLUTION :

Given equation is $\frac{sin2A}{sinA}-\frac{cos2A}{cosA}=secA$

Now taking the LHS :

$\frac{sin2A}{sinA}-\frac{cos2A}{cosA}$

By using the trignometric formulae :

i) $sin2x=2sinxcosx$

ii) $cos2x=1-2sin^2x$

$=\frac{2sinAcosA}{sinA}-\frac{(1-2sin^2A)}{cosA}$

$=2cosA-\frac{(1-2sin^2A)}{cosA}$

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

$=\frac{2(cos^2A+sin^2A)-1}{cosA}$

By using the trignometric formula :

$cos^2x+sin^2x=1$

$=\frac{2(1)-1}{cosA}$

$=\frac{1}{cosA}$

By using the trignometric formula :

$\frac{1}{cosx}=secx$

$=secA$ = RHS

⇒ LHS = RHS

∴ $\frac{sin2A}{sinA}-\frac{cos2A}{cosA}=secA$

∴ the given equation  $\frac{sin2A}{sinA}-\frac{cos2A}{cosA}=secA$ is true and verified.