verify the identity. \n\n$(sec x+\tan x)^{2}=\frac{1+sin x}{1-sin x}$\n\nto verify the identity, start with…

verify the identity. \n\n$(sec x+\tan x)^{2}=\frac{1+sin x}{1-sin x}$\n\nto verify the identity, start with the more complicated side and transform it to look like the other side. choose the correct transformations and transform the expression at each step. \n\n$(sec x+\tan x)^{2}$\n$=square$\n(rewrite in terms of sine and cosine inside parentheses. \n(do not simplify.)
Answer
Explanation:
Step1: Use trigonometric identities
Recall that (\sec x=\frac{1}{\cos x}) and (\tan x = \frac{\sin x}{\cos x}). [ \left(\sec x+\tan x\right)^{2}=\left(\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right)^{2} ]
Answer:
[ \left(\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right)^{2} ]