prove: 1/(1-sinX) + 1/(1+sinX)= 2sec^2 X?

Question

anyone got any solutions here PLEASE

Answer 1

1/(1 - sin x) + 1/(1 + sin x) = 2sec^2 x
((1 + sin x) + (1 - sin x))/((1 - sin x)(1 + sin x)) = 2sec^2 x
2/(1 - sin^2 x) = 2sec^2 x
2/(cos^2 x) = 2sec^2 x
2sec^2 x = 2sec^2 x

Answer 2

haha... trig proofs... ok here we go.

multiply to get common denominators, and in this case, we would use the conjugates. so after the first step, you would have

(1 + sinX)/ (1 - sin^2 X) + (1 - sinX)/(1 - sin^2 X)

then simplify you'd get:
2 / (1 - sin^2 X)

now remember the identity sin^2 X + cos^2 X = 1
thus sin^2 X = 1 - cos^2 X
and plug that into the denominator and you get:

2 / (1 - (1 - cos^2 X))
=
2 / (1 - 1 + cos^2 X)
=
2 / (cos^2X)

now, 1/cosX = secX then 1/ cos^2 X = sec^2 X and substitute

2 * ( 1 / (cos^2 X))
=
2 * sec^2 X
=
2sec^2 X


some of those steps you may leave out as they can be done in your head..but i showed them just for the heck of makign it clear

Answer 3

1/(1-sinx) + 1/(1+sinx) = 2sec²x

((1+sinx) +(1-sinx))/(1 - sin²x)= 2sec²x

(2)/(1 - sin²x) = 2sec²x

(2)/(cos²x) = 2sec²x

2sec²x = 2sec²x
.