Question

Simplify the trigonometric expression
1/1+sin0 + 1/1-sin0

Answer 1

Use parentheses :P the denominator needs them

Answer 2

\[\frac{ 1 }{ 1+\sin \theta }+ \frac{ 1 }{ 1-\sin }\]

Answer 3

Wow good job.

Answer 4

Get common denominators:\[\large \frac{ 1(1-\sin \theta) }{ (1+\sin \theta) (1-\sin \theta) }+ \frac{ 1 (1+\sin \theta)}{ (1-\sin \theta) (1+\sin \theta) }\]

Answer 5

\[\frac{ 1}{ 1+\sin \theta }+\frac{ 1}{ 1-\sin \theta }\]

Answer 6

thats what i meant ^

Answer 7

the answer is 2 right?

Answer 8

you need a common denominator here. Multiply the left expression by (1-sintheta/1-sintheta), and multiply the expression on the right by (1+sin theta/1+ sin theta), like agent0smith did.

Answer 9

\[\large \frac{ 1(1-\sin \theta) }{ (1+\sin \theta) (1-\sin \theta) }+ \frac{ 1 (1+\sin \theta)}{ (1-\sin \theta) (1+\sin \theta) } =\]\[\large \frac{2}{ (1-\sin \theta) (1+\sin \theta) } = \frac{2}{ 1-\sin^2 \theta}\]

Answer 10

There is an identity hidden in the 1 - sin^2 theta...

Answer 11

1 - sin^2 theta = cos^2 theta.

Answer 12

\[\frac{ 2 }{ \cos ^{2} }\]

Answer 13

Oh yeah \[\large \frac{2}{ 1-\sin^2 \theta} = \frac{ 2 }{ \cos^2 \theta }= 2 \sec ^2 \theta\]

Answer 14

Cool! This just goes on and on.... ; )

Answer 15

thank you