Here's the question you clicked on:
starkeyfish
Verify the identity: (1-sinx)/cosx = cosx/(1+sinx)
LS = (1-sinx)/cosx = 1/cosx - sinx/cosx RS = \(\large\frac{cosx}{1+sinx}\) = \(\large\frac{cosx(1-sinx)}{(1+sinx)(1-sinx)}\) = \( \large\frac{cosx}{1-sin^2x} - \frac{cosxsinx}{1-sin^2x}\) =.... Note: \[1 = cos^2x + sin^2x \]
I'm very bad at basic math , so I'm finding all of this to be very foreign.
Actually, simply working from the left is ok :D \[RS = \frac{cosx}{(1+sinx)}\]\[=\frac{cos}{1+sinx}\times \frac{1-sinx}{1-sinx}\]By the identity a^2 - b^2 = (a+b)(a-b), we get\[=\frac{cos(1-sinx)}{1-sin^2x}\]Then, \(sin^2x+cos^2x =1\) to express \(1-sin^2x\) in terms of \(cos^2x\). Finally, cancel the common factor, and you'll get the left side!
so 1-sin^2 = cos^2x?