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aggy113 · Sep 24, 2009, 03:30 PM

prove the following identities?
a.) (1 - cosX)(cscX + cotX)=sinX

ArcSine · Sep 25, 2009, 03:04 AM

Here's one way your proof could proceed: Working on the left-hand side, multiply it out:

$csc\theta + cot\theta - cos\theta csc\theta - cos\theta cot\theta$

Re-express the terms (I'll let you fill in the steps in your proof):

$\frac{1}{sin\theta}\ +\ \frac{cos\theta}{sin\theta} \ - \ \frac{cos\theta}{sin\theta} \ - \ \frac{cos^2 \theta}{sin\theta}$

Note that two of the terms cancel out, and the remaining two terms, having a common denominator, can be written

$\frac{1-cos^2 \theta}{sin\theta}$

Now look at the numerator, and think of the Pythagorean Identity. Can you finish it up from here?

Also keep in mind that with many trig IDs there are multiple possible paths to the same conclusion. Good luck!

TANDE005 · Jan 12, 2010, 06:10 PM

How do I solve the following identity?

secxcscx=tanx + cotx

galactus · Jan 12, 2010, 07:38 PM

Please start your own thread.

But here goes:

Start with the right side and turn it into the left side.

$tan(x)+cot(x)=\frac{sin(x)}{cos(x)}+\frac{cos(x)}{ sin(x)}$

Cross multiply:

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

Notice the famous identity in the numerator is equal to 1.

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

$=sec(x)csc(x)$