sinabsinabsin2asin2bSolution1 1secx tanx 1secx tanx 2tanx

sin(a+b)sin(a-b)=sin^2a-sin^2b

Solution

1) 1/(secx -tanx) - 1/(secx +tanx) = 2tanx

LHS :taking (secx -tanx)(secx +tanx) as lcd

[ secx +tanx - secx +tanx]/(secx -tanx)(secx +tanx)   

=2tanx/(sex^2x -tan^2x)

as per identity 1+tan^2x = sec^2x

So, 2tanx/(1) = 2tan x

RHS

2) (1+ tan^2x)cos^2x =1

LHS

cos^2x +tan^2xcos^2x

cos^2x + (sin^2x/cos^2x)cos^2x

= cos^2x +sin^2x

= 1

4) sin(A+B)sin(A-B) = sin^2A - sin^2B

LHS: sin(A+B)sin(A-B)

(sinAcosB +cosAsinB)(sinAcosB -cosAsinB)

=(sinAcosB)^2 -(cosAsinB)^2

As per identity : 2 sin A sin B = cos(A B) cos(A + B)

So, [cos(A+B - (A-B)) - cos(A+B+A -B)]/2

= [cos(2B ) - cos(2A)]/2

= [(1-2sin^2B ) - (1-2sin^2A)]/2

= sin^2A - sin^2B

RHS