Question

Please help prove the identity of (1-sinx)/(cosx)=(1)/(secx+tanx)

Answer 1

It's easier if we work from the right side, and make it match the left.
If you need to do it from the other direction, then just follow these steps in reverse.

\[\large \frac{1}{\sec x + \tan x}\]Working with this side, we'll start by converting everything to sines and cosines.\[\large \frac{1}{\dfrac{1}{\cos x}+\dfrac{\sin x}{\cos x}} \qquad = \qquad \frac{1}{\left(\dfrac{1+\sin x}{\cos x}\right)}\]
Since we're dividing by a fraction, we can rewrite it as multiplication by flipping the bottom fraction.\[\large \frac{\cos x}{1+\sin x}\]

Answer 2

From here, we'll multiply the top and bottom by (1-sin x).\[\large \frac{\cos x}{1+\sin x}\left(\frac{1-\sin x}{1-\sin x}\right) \qquad = \qquad \frac{\cos x(1-\sin x)}{1-\sin^2x}\]

Answer 3

Going back to our most basic trig identity:\[\large \color{cornflowerblue}{\sin^2x+\cos^2x=1} \qquad \rightarrow \qquad \color{cornflowerblue}{\cos^2x=1-\sin^2x}\]Applying this to our problem gives us,\[\large \frac{\cos x(1-\sin x)}{\color{cornflowerblue}{1-\sin^2x}} \qquad = \qquad \frac{\cos x(1-\sin x)}{\color{cornflowerblue}{\cos^2x}}\]

Answer 4

Then we can cancel out a cosine from the top and bottom and we're done! Yay!\[\large \frac{\cancel{\cos x}(1-\sin x)}{\cos^{\cancel{2}}x} \qquad = \qquad \frac{1-\sin x}{\cos x}\]