Question

I have two questions, both are verifying trigonometric identities, they are:

(secx - tanx)^2 + 1
---------------------- = 2tanx
csc(secx-tanx)

and

1 + cosx + sinx
--------------------- = secx + tanx
1 + cosx - sinx


please show all the steps to reach these two identities, and thank you so much for your help.

Answer 1

1. $\frac{(secx - \tan x)^2 + 1}{cscx(secx - \tan x)} = 2\tan x,$

$$s e c x = \frac {1}{c o s x}, c s c x = \frac {1}{s i n x}, t a n x = \frac {s i n x}{c o s x}, c o s ^ {2} x + s i n ^ {2} x = 1$$ $$\begin{array}{l} \frac {(s e c x - t a n x) ^ {2} + 1}{c s c x (s e c x - t a n x)} = \frac {\left(\frac {1}{c o s x} - \frac {s i n x}{c o s x}\right) ^ {2} + 1}{\frac {1}{s i n x} \left(\frac {1}{c o s x} - \frac {s i n x}{c o s x}\right)} = \frac {\frac {1}{c o s ^ {2} x} ((1 - s i n x) ^ {2} + c o s ^ {2} x)}{\frac {1}{s i n x} \left(\frac {1}{c o s x} - \frac {s i n x}{c o s x}\right)} = \\ = \frac {\frac {s i n x}{c o s ^ {2} x} ((1 - s i n x) ^ {2} + c o s ^ {2} x)}{\frac {1}{c o s x} (1 - s i n x)} = \frac {\frac {s i n x c o s x}{c o s ^ {2} x} ((1 - s i n x) ^ {2} + c o s ^ {2} x)}{(1 - s i n x)} = \\ \frac {\frac {s i n x}{c o s x} ((1 - s i n x) ^ {2} + c o s ^ {2} x)}{(1 - s i n x)} = \frac {\frac {s i n x}{c o s x} (1 - 2 s i n x + s i n ^ {2} x + c o s ^ {2} x)}{(1 - s i n x)} = \\ = \frac {\frac {s i n x}{c o s x} (1 - 2 s i n x + 1)}{(1 - s i n x)} = \frac {\frac {2 s i n x}{c o s x} (1 - s i n x)}{(1 - s i n x)} = \frac {2 s i n x}{c o s x} = 2 t a n x. \\ \end{array}$$

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

$$\begin{array}{l} \frac {1 + \cos x + \sin x}{1 + \cos x - \sin x} = \frac {1 + \cos x + \sin x}{1 + \cos x - \sin x} \times \frac {1 + \cos x - \sin x}{1 + \cos x - \sin x} = \\ = \frac {1 + \cos x - \sin x + \cos x + \cos^ {2} x - \sin x \cos x + \sin x + \sin x \cos x - \sin^ {2} x}{1 + \cos x - \sin x + \cos x + \cos^ {2} x - \sin x \cos x - \sin x - \sin x \cos x + \sin^ {2} x} = \\ = \frac {1 + 2 \cos x + \cos^ {2} x - \sin^ {2} x}{1 + 2 \cos x - 2 \sin x - 2 \sin x \cos x + 1} = \frac {\cos^ {2} x + \sin^ {2} x + 2 \cos x + \cos^ {2} x - \sin^ {2} x}{2 + 2 \cos x - 2 \sin x - 2 \sin x \cos x} = \\ = \frac {2 \cos^ {2} x + 2 \cos x}{2 ((1 + \cos x) - \sin x (1 + \cos x))} = \\ = \frac {2 \cos x (\cos x + 1)}{2 (1 + \cos x) (1 - \sin x)} = \frac {\cos x}{1 - \sin x} \times \frac {1 + \sin x}{1 + \sin x} = \frac {\cos x (1 + \sin x)}{1 - \sin^ {2} x} = \frac {\cos x (1 + \sin x)}{\cos^ {2} x} = \\ = \frac {1 + \sin x}{\cos x} = \frac {1}{\cos x} + \frac {\sin x}{\cos x} = \sec x + \tan x. \\ \end{array}$$