Question

I need help with this practice problem solving I could not get the entire problem in one photo so I will send an additional photo of the rest

Answer 1

Answer:

$\frac{\cos x}{1-\sin x}=\frac{1}{\cos x}+\frac{\sin x}{\cos x}$$\frac{\cos x}{1-\sin x}=\frac{1+\sin x}{\cos x}$$\frac{\cos x}{1-\sin x}=\frac{(1+\sin x)\cdot\cos x}{1-\sin ^2x}$$\frac{\cos x}{1-\sin x}=\frac{(1+\sin x)\cdot\cos x}{(1-\sin^{}x)(1+\sin x)}$$\frac{\cos x}{1-\sin x}=\frac{\cos x}{1-\sin x}$

Explanation:

First, we realise that

$\sec x=\frac{1}{\cos x}$

and

$\tan x=\frac{\sin x}{\cos x}$

therefore,

$\sec x+\tan x=\frac{1}{\cos x}+\frac{\sin x}{\cos x}$

Now,

$\frac{1}{\cos x}+\frac{\sin x}{\cos x}=\frac{1+\sin x}{\cos x}$

Using the fact that

$\frac{1}{\cos x}=\frac{\cos x}{\cos^2x}=\frac{\cos x}{1-\sin^2x}$

Therefore,

$\frac{1+\sin x}{\cos x}=\frac{\cos x(1+\sin x)}{1-\sin^2x}$$=\frac{\cos x(1+\sin x)}{(1-\sin x)(1+\sin x)}=\frac{\cos x}{1-\sin x}$

Hence, we have shown that

$\boxed{\sec x+\tan x=\frac{\cos x}{1-\sin x}\text{.}}$