Question

prove that √(1−sinA/1+sinA) = secA tanA.

Answer 1

Answer:

The given statement is proved

Answer 2

Answer:

sry I got secA-tanA

Step-by-step explanation:

first consider cosA=√(sin^2A)

=√((1+sinA)(1-sinA))

therefore √1+sinA=cosA/√(1-sinA)

and √1-sinA=cosA/√(1+sinA)

and also, 1+sinA=cos^2A/1-sinA

√1-sinA/√1+sinA

=(cosA/√1+sinA)/√1+sinA

=cosA/1+sinA

substituting values

=cosA(1-sinA)/cos^2A

=1-sinA/cosA

=(1/cosA)-(sinA/cosA)

=secA-tanA