Question

Verify each identity:
(sinx-cosx)/(sinx) + (cosx-sinx)/(cosx)=2-secxcscx

Answer 1

This looks fun. Any ideas on how to start?

Answer 2

common denominator?

Answer 3

Yes

Answer 4

And what do you get?

Answer 5

\[\frac{ \cos \theta \sin \theta-\cos^2 \theta +\sin \theta \cos \theta -\sin^2 \theta }{ \sin \theta \cos \theta }\]

Answer 6

Correct, now combine like terms, and separate the fraction into parts.

Answer 7

\[\frac{ 2\cos \theta \sin \theta }{ \cos \theta \sin \theta} + \frac{ -\cos^2 \theta - \sin^2 \theta }{ \cos \theta \sin \theta }\]

Answer 8

There ya go, now simplify. Take out the negative and make it:
\[\LARGE -(\frac{sin^2\theta+cos^2\theta}{cos\theta sin\theta})\]
Do you remember what \[\LARGE sin^2+cos^2x=?\]

Answer 9

1

Answer 10

Right. In the first fraction, the cos and sin cancel leaving 2.
The sin+cos turn to 1. Now we separate that:
\[\LARGE 2+(\frac{1}{cos\theta}*\frac{1}{sin\theta})\]
Finish it off~

Answer 11

\[2-\sec \theta \csc \theta = 2- \sec \theta \csc \theta\]

Answer 12

Whoops, forgot my negative, but yea :)

Answer 13

Thank you so much!!!!! :DDDDDDDDDDD

Answer 14

Anytime~