Question

CosA/(1+sinA)+(1+sinA)/cosA=2secA

Answer 1

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

L.H.S

cosA/(1+sinA) + (1+sinA)/cosA = [ cos2A + (1+sinA)2 ] / (1+sinA)cosA (tking lcm)
-------> = cos2A + 1+ sin2A + 2sin A / (1+sinA)cosA
-------> = (cos2A + sin2A) +1+ 2sin A / (1+sinA)cosA
since cos2A+ sin2A = 1 ----- > 1+ 1+ 2sin A / (1+sinA)cosA
-------> 2+ 2sin A / (1+sinA)cosA
-------> 2(1 + sinA) / (1+sinA)cosA
-------> cancel the common term (1+sinA)
-------> 2/cosA = 2secA = R.H.S ( 1/cosA=secA)

Answer 2

Answer:

Step-by-step explanation: