Question

1+cosA+sinA/1+cosA-sinA=1+sinA/cosA

Answer 1

Answer:

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

L.H.S.

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

Multiplying by [sinA+(1-cosA)] in Nr and Dr.

=[sinA+(1-cosA)]^2/[sin^2A-(1-cosA)^2]

=[sin^2A+2.sinA.(1-cosA)+(1-cosA)^2]/[(1-cos^2A)-(1-cosA)^2]

=[(1-cos^2A)+2.sinA.(1-cosA) +(1-cosA)^2]/[(1-cosA)(1+cosA) - (1-cosA)^2].

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

=[2+2sinA]/[2.cosA].

=2(1+sinA)/2.cosA.

= (1+sinA)/cosA , proved.

Answer 2

LHS

$\frac{1 + cosa + sina}{1 + cosa - sina} \\ \\ = \large \frac{ \frac{1 + cosa + sina}{cosa} }{ \frac{1 + cosa - sina}{cosa} } \\ \\ = \frac{seca + tana + 1}{seca - tana + 1} \\ \\ = \frac{(seca + tana) + ( {sec}^{2}a - {tan}^{2}a) }{seca - tana + 1} \\ \\ = \frac{(seca + tana)(1 + seca - tana)}{seca - tana + 1} \\ \\ = seca + tana \\ \\ = \frac{1 + sina}{cosa}$

RHS