Question

1+sin2x/1-sin2x=tansguar{π/4+x

Answer 1

I don't know. I am sorry

Answer 2

Answer:

Step-by-step explanation:

LHS=√1+sin2x/√1-sin2x

=√(sinx+cosx)^2/√(sinx-cosx)^2. [Using=1+sin2x=(sinx+cosx)^2,1-sin2x=(sinx-cosx)^2]

√(cosx+sinx)^2/√(cosx-sinx)^2

[Now divide numerator and denominator by cosx]

=1+tanx/1-tanx = tanπ/4+tanx/1-tanx. Tanπ/4. [ By using tanx+tany/1-tanx. Tany]

Tan(π/4+x)

LHS=RHS