please prove this trigonomeric identity: 1/(1+sin) + 1/(1-sin) = 2sec^2?

Answer 1

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

Answer 2

Take LHS
1. cross multiply the terms
that is 1 - sin + 1 + sin/ (1-sin)(1+sin)
2 . 2/(1-sin^2) (a+b)(a-b)=a^2+b^2
3. 2/cos^2
4. 2sec^2
Thus LHS = RHS
so easy mate.just try

Answer 3

1/(1 + sinx) + 1/(1 - sinx) = 2sec^2(x)

[Note: remember to include your variable inside each trig function; otherwise the question doesn't make sense!]

We choose the left hand side (LHS) because it is more complex.

LHS = 1/(1 + sinx) + 1/(1 - sinx)

Put them both under a common denominator.

(1 - sinx) /[(1 + sinx) (1 - sinx)] + (1 + sinx) /[(1 + sinx) (1 - sinx)]

Merge into one fraction.

([1 - sinx] + [1 + sinx])/[(1 + sinx) (1 - sinx)]

Simplify the numerator.

[2]/[(1 + sinx) (1 - sinx)]

Note that the denominator is how we factor difference of squares. For that reason, it will become a difference of squares. Simplify the denominator.

[2]/[1 - sin^2(x)]

Note that 1 - sin^2(x) = cos^2(x), so we'd get

[2]/(cos^2(x))

We can change this to

(2) [1/[cos^2(x)]]

And by definition, this is equal to

2sec^2(x) = RHS

Answer 4

1/(1+sin) + 1/(1-sin) =
(1-sin)/(1-sin^2) + (1+sin)/(1-sin^2) =
[(1-sin +1+sin)]/(1-sin^2) =
2/(1-sin^2) =
2/(cos^2) =
2sec^2 = 2sec^2
tadah.

Answer 5

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