Question

Prove the identity. (show steps please!) Thank you!

tan^2(x/2+π/4)=(1 + sin x)/(1 − sin x)

Answer 1

Recall that formula for sum in tangent:
tan (a +/- b) = [tan(a) +/- tan(b)] / [1 -/+ tan(a)tan(b)]
Applying this,
tan^2(x/2 + π/4)
= { [tan(x/2) + tan(π/4)] / [1 - tan(x/2)tan(π/4)] }^2
Note that tan(π/4) = 1, so
= [(tan(x/2) + 1) / (1 - tan(x/2))]^2
= (tan^2 (x/2) + 2*tan(x/2) + 1) / (1 - 2*tan(x/2) + tan^2 (x/2))
Note that 1 + tan^2 (x/2) = sec^2 (x/2). Thus,
= (sec^2 (x/2) + 2*tan(x/2)) / (sec^2 (x/2) - 2*tan(x/2))
Note that sec (x/2) is also 1 / cos(x/2), and tan (x/2) = sin(x/2) / cos(x/2):
= (1/cos^2 (x/2) + 2sin(x/2)/cos(x/2)) / (1/cos^2 (x/2) - 2sin(x/2)/cos(x/2))
= (1 + 2sin(x/2)cos(x/2))/(cos^2 (x/2)) / (1 - 2sin(x/2)cos(x/2))/(cos^2 (x/2))
The cos^2 (x/2) will be cancelled.
= (1 + 2sin(x/2)cos(x/2)) / (1 - 2sin(x/2)cos(x/2))
Finally, recall that 2sin(a)cos(a) = sin(2a). Thus if a is actually 1/2x, it will become sin(2*(1/2)x) or sin(x). Rewriting:
= (1 + sin(x)) / (1 - sin(x))

Hope this helps :D