Question

prove that

and plz friends I need it very urgent​

Answer 1

Step-by-step explanation:

LHS

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

=√[{(1+sinA)-cosA}/{(1+sinA)+cosA)}]²

=√[(1+sinA)²-2(1+sinA)cosA+cos²A/ (1+sinA)²+2(1+sinA)cosA+cos²A]

=√1+2sinA+sin²A-2cosA-2sinAcosA +cos²A/1+2sinA+sin²A+2cosA+2sinAcosA

+cos²A

=√1+sinA-cosA(1+sinA)/1+sinA+2cosA(1+sinA)

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

=√(1+sinA)(1-cosA)/(1+sinA)(1+cosA)

=√1-cosA/1+cosA

thank You