Question

Verify the identity:
(cos x)/(1 + sin x) + (1 + sin x)/(cos x) = 2sec x

My work:
((cos x)cos x)/(1 - sin x)(cos x) + ((1 - sin x))(1 + sin x))/(cos x)(1 - sin x) = 2 sec x

((cos x)^2 / cos x - sin x cos x) + (1 - sin^2 x)/(cos x - sin x cos x) = 2 sec x

(cos^2 x + 1 - sin^2 x)/(cos x - sin x cos x) = 2 sec x
(2 cos^2 x) / cos x - sin x cos x) = 2 sec x

Is there something I'm missing off the final equation or did I mess up earlier?

Answer 1

A couple of signs are incorrect. I'd get to: (2 cos^2 x) / cos x + sin x cos x) = 2 sec x

Answer 2

is it (1+sinx) or (1-sinx) it changes after 1st line ?

Answer 3

((cos x)cos x)/(1 + sin x)(cos x) + ((1 + sin x))(1 + sin x))/(cos x)(1 + sin x) = 2 sec x

((cos x)^2 / cos x + sin x cos x) + (1 + sin x)^2/(cos x + sin x cos x) = 2 sec x

((cos x)^2 / cos x + sin x cos x) + (1 + sin x)^2/(cos x + sin x cos x) = 2 sec x
((sin x)(cos x))^2 / (cos x + sin x cos x)

Sorry new work right here.

Answer 4

((sin x)(cos x))^2 / (cos x + sin x cos x) = 2 sec x

Answer 5

Sorry I'm out of it this evening.

Answer 6

ok something is off on last line
\[\cos^{2} x + (1+\sin x)^{2} = \cos^{2}x + \sin^{2} x +2\sin x +1 = 2+2\sin x = 2(1+\sin x)\]

Answer 7

how are you getting
\[\sin^{2}x \cos^{2} x\]

Answer 8

((1 + sin x)(cos x))^2 / (cos x + sin x cos x) = 2 sec x

Answer 9

So then you get:
cos^2x+sin^2x+2sinx+1
from distributing on the numerator, then
2+2sinx
because cos^2x + sin^2 x = 1
Take it back out:
2(1 + sin x)/(cos x + sin x cos x = 2 sec x

I understand this. divide by two to rid of those coefficients:

(1 + sin x)/(cos x + sin x cos x) = sec x

Then what?

Answer 10

wait try not to change right hand side when verifying
\[\frac{2(1+\sin x)}{\cos x (1+\sin x)} = \frac{2}{\cos x} = 2 \sec x\]

Answer 11

Well I've been told as long as it's an easily understandable change, then it's okay to make it easier, but I understand why it should be kept the same while verifying.

2(1+sinx)/cosx(1+sinx)=2/cosx=2secx

Okay so I just forgot about factoring out a cos in the denominator. Thank you very much!

Answer 12

your welcome