Question

Prove that :- tan (π/4+x/2)=1/secx-tanx

Answer 1

LHS = tan(π/4 + x/2)

we know, tan(A + B) = (tanA+ tanB)/(1 - tanA.tanB)

= (tanπ/4 + tanx/2)/(1 - tanπ/4. tanx/2)

we know, tanπ/4 = 1

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

= {1 + sin(x/2)/cos(x/2)}/{1 - sin(x/2)/cos(x/2)}

= {cos(x/2) + sin(x/2)}/{cos(x/2) - sin(x/2)}

= (cosx/2 + sinx/2)(cosx/2 + sinx/2)/(cosx/2 - sinx/2)(cosx/2 + sinx/2)

= (cosx/2 + sinx/2)²/(cos²x/2 - sin²x/2)

cos²A - sin²A = cos2A
2sinA.cosA = sin2A

= (cos²x/2 + sin²x/2 + 2cosx/2.sinx/2)/cosx

= (1 + sinx)/cosx

= secx + tanx

= (secx + tanx)/(sec²x - tan²x)

as sec²A - tan²A = 1

= 1/(secx - tanx) = RHS

Answer 2

Refer both the pages posted.

I hope this answer helps you .