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_chanelxo_ · Mar 5, 2013, 02:11 PM

Verifying trigonometric identities

_chanelxo_ · Mar 5, 2013, 02:14 PM

PLEASE SOLVE sin x + cos x /sin x - cos x-sin x/ cos x = sec x csc x

ebaines · Mar 6, 2013, 10:25 AM

Originally Posted by _chanelxo_

PLEASE SOLVE sin x + cos x /sin x - cos x-sin x/ cos x = sec x csc x

Please use parentheses. What you wrote is this:

$<br /> \sin x + \frac {\cos x}{\sin x} - \cos x - \frac {\sin x}{\cos x} = secx\ csc x<br />$

which is incorrect. I believe what you meant is this:

$<br /> \frac {(\sin x + \cos x)}{\sin x} \ - \ \frac {(\cos x - \sin x)}{\cos x} = sec x \ csc x<br />$

As usual for working with trig identities start by converting any tanx, cotx, secx, and cscx functions to their sinx and cosx equivalents. Then it's a simple matter of combining the two fractions, simplifying, and applying the identity:

$\sin ^2 x + \cos^2x = 1$