Question

I need help proving trigonometric identities[(1+sinθ)÷cosθ]+[cosθ÷(1+sinθ)] is identical to 2secθ

Answer 1

'Ill use s for sin and c for cos

(1+s) / c + c / (1+x)

= [(1+s)(1+s) + c^2] / c(1+s)

= 1 + 2s + s^2 + c^2 / c(1+s)

= 2+2s / c(1+s) (because s^2 + c^2 = 1}

= 2(1+s) / c(1+s)

= 2/c

= 2 / cos θ

= 2 sec θ answer

Answer 2

`((1+sin)/(1 + sin))(1 + sin)/(cos) + ((cos)/(cos) )(cos)/(1 + sin) = 2 sec`

`(1 + 2sin + sin^(2) )/(cos(1 + sin)) + (cos^(2) )/(cos(1 + sin)) = 2 sec`

`(1 + 2sin + sin^(2) + cos^(2))/(cos(1 + sin)) = 2 sec`

`(1 + 2sin + 1)/(cos(1 + sin)) = 2 sec`

`(2 + 2sin)/(cos(1 + sin)) = 2 sec`

`(2(1 + sin))/(cos(1 + sin)) = 2 sec`

`(2)/(cos) = 2 sec`

2 secθ = 2 secθ