Question

cosA/1+sinA + 1+sinA/cosA = 2/cosA = 2secA proved that.​

Answer 1

$\huge\red{\underline{{\boxed{\textbf{QUESTION}}}}}$

Prove:-

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

$\huge\red{\underline{{\boxed{\textbf{ANSWER}}}}}$

L.H.S

cosA/(1+sinA) + (1+sinA)/cosA

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

@HarshPratapSingh

Answer 2

Answer:

L.H.S

cosA/(1+sinA) + (1+sinA)/cosA

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