Trig Identities

**Giggly2** · April 30th 2009, 04:15 PM

I'm stuck on these trig identities and I was wondering if anyone else could figure them out?? The first time I typed them in they were kinda hard to understand so I hope I did a better job this time.

(secX/sinX) - (2sinX/cosX) = cotX - tanX

(tanX-sinX) / sin(cubed)X = secx / (1 + cosX)

(cosX + 1)/(secX -tanX) - (1-cosX)/(secX + tanX) = 2(1+tanX)

1 + 3sin(squared)Xsec(to the 4th)X = sec(to the 6th)X - tan(to the 6th)X

1 + (2/tanX+cotX) + (1/sec(squared)X +csc(squared)X) = (1 + sinXcosX)(squared)

**running-gag** · May 1st 2009, 12:05 AM

Hi

Here is the first one

$\frac{\sec x}{\sin x} - 2 \frac{\sin x}{\cos x} = \frac{1}{\cos x \sin x} - 2 \frac{\sin^2 x}{\cos x \sin x} = \frac{1-2\sin^2 x}{\cos x \sin x} = \frac{1-\sin^2 x-\sin^2 x}{\cos x \sin x}$

$\frac{\sec x}{\sin x} - 2 \frac{\sin x}{\cos x} = \frac{\cos^2 x-\sin^2 x}{\cos x \sin x} = \frac{\cos^2 x}{\cos x \sin x} - \frac{\sin^2 x}{\cos x \sin x} = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}$

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