+0

# Trigonometry identities

0
31
3

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

Well here is my steps:

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

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

What to do from here?

Jan 30, 2020

#1
+18252
0

On the left-hand side, when you multiplied [ 1 + sin(x) ]  times another [ 1 + sin(x) ],

you should get:     [ 1 + sin(x) ]2  =  1 + 2sin(x) + sin2(x)

Then continue ...

Jan 30, 2020
#2
0

Hello geno,

I continued this step but didn't write it

I.e. here it is:

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

$$sin^2(x)+cos^2(x)=1$$

Thus we have:

$$1+1+2sin(2)/cos(x)(1+sin(x))$$

Which is

$$\frac{2+2sin(x)}{cos(x)(1+sin(x))}$$ So the required is this should be simplified to be 2/cos(x)

Guest Jan 30, 2020
#3
0

nvm solved it, just factor 2 out of the numerator and 1+sin(x) cancels left with 2/cos(x)

Guest Jan 31, 2020