Question

cosx/ (1+sinx) + (1+sinx)/ cosx= 2secx

Answer 1

\[\large \frac{cos(x)}{1 + sin(x)} + \frac{1 + sin(x)}{cos(x)} = 2sec(x)\]

We need a common denominator first...what would that be?

Answer 2

We can see the first fraction is missing a factor of cos(x) on the bottom...and we can see that the second fraction is missing a factor of 1 + sin(x) so we can work with that...

\[\large \frac{cos(x)}{cos(x)} \times \frac{cos(x)}{1 + sin(x)} + \frac{1 + sin(x)}{1 + sin(x)} \times \frac{1 + sin(x)}{cos(x)} = 2sec(x)\]

And simplify that down a bit

\[\large \frac{cos^2(x)}{(1 + sin(x))cos(x)} + \frac{(1 + sin(x))^2}{(1 + sin(x))cos(x)} = 2sec(x)\]

Now lets put them over the common denominator

\[\large \frac{cos^2(x) + (1 + sin(x))^2}{(1 + sin(x))cos(x)} = 2sec(x)\]

Now...what can we replace \(\large cos^2(x)\) with?