[shant112]

can someone help me solve this problem please

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

[kyop]

can someone help me solve this problem please

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

I like this problem!

1/(1-sinx) + 1/(1+sinx) = 2sec^2(x) can be rewritten as:

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

Now, multiply the whole equation by (1-sinx)(1+sinx)

Significant cross-cancelling will give you:

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

Notice that the sinx and -sinx cancel out, leaving you:

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

Now, if you FOIL out the binomials, you get 2(1-sin^2x)/cos^2x

(1-sin^2x) actually equals cos^2x so you end up with:

2 cos^2x /cos^2x which just equals 2.

So, when all is said and done, you have 1+1 = 2 which is always true.

Now, you have a couple of times when x CANNOT equal something. For instance, x cannot equal 90 degrees (pi/2 radians) because sinx would equal -1, causing 1/0 in our original equation. Same with x = 270 (3pi/2 radians).

So, your solution is all reals except 90 and 270.

[kyop]

Good point with the coterminal points. I just generally work in the 0 to 2pi range unless otherwise stated.