Question

PROVE THE ABOVE EQUATION.
AND PLEASE ANSWER PROPERLY!!

If u don't know the answer then just leave it and don't put unwanted statements..........
please...........​​

Answer 1

Answer:

cos(90° - A) = sinA

sin(90° - A) = cosA

the LHS becomes

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

(sin^2A + (1 + cos^2A))/(sinA(1+cosA))

(sin^2A + 1 + cos^2A + 2 cosA)/(sinA(1+cosA))

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

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

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

2/sinA

2cosecA

RHS

Answer 2

cos(90-A)=sinA
sin(90-A)=cosA
So
sinA/1+cosA + 1+cosA/sinA
Now we take the LCM
sin^2A+(1+cosA)^2/sinA(1+cosA)
We have the formula sin^2A+cos^2A=1
sin^2A+1+cos^2A+2cosA/sinA(1+cosA)
1+1+2cosA/sinA(1+cosA)
2+2cosA/sinA(1+cosA)
Take 2common
2(1+cosA)/sinA(1+cosA)
1+cosA gets canceled
2/sinA
1/sinA=cosecA
=2cosecA (proved)