# how to prove the following trig identities?

1. $\frac{\sin^2 x + \cos^2 x}{\csc x} = \sin x$
2. $(1 - \tan x)^2 = \sec^2 x - 2 \tan x$
3. $\tan^2 x - \sin^2 x = \tan^2 x \sin^2 x$
4. $\frac{\cos^2 x - 1}{\cos x} = -\tan x \sin x$
5. $\cos x (\tan x + \cot x)= \csc x$
6. $\tan x \cot x - \cos^2 x = \sin^2 x$
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Yeesh. Just replace all tan, cot, csc and sec functions with pure sine's and cosines according to their definitions. Then it's just algebraic manipulation and $\sin^2+\cos^2=1$ all the way down. –  anon Nov 19 '11 at 0:29
As anon said, replace $\csc x=1/\sin x$, $\sec x=1/\cos x$, $\tan x=\sin x/\cos x$, and $\cot x = \cos x/\sin x$. With everything written as sines and cosines the identities will pop right out -- well, ok -- You'll also need the identity: $\sin^2 x+\cos^2 x =1$ (and it's variants like $\sin^2 x=1-\cos^2 x$ etc). –  Bill Cook Nov 19 '11 at 0:37
tcomer, you should try and make an effort to explain exactly what you tried and where you got stuck, when asking this kind of homework questions (even if these are not actually your homework) –  Mariano Suárez-Alvarez Nov 19 '11 at 2:19

1. $\sin^2 x+\cos^2 x=1$. sine is the reciprocal of cosecant.

2. Clear the parentheses, then use $1+ \tan^2x=\sec^2x$.

3. Factor out $\sin^2x$ and simplify.

4. Factor out $-1$ from the numerator and the rest should be easy.

5. This looks to be the tough one. Try showing that $\sin x \cos x( \tan x+ \cot x)=1$, then divide both sides by $\sin x$.

6. Tangent and cotangent are reciprocals.

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That's awfully generous, considering the miniscule amount of effort put forth by the OP... –  The Chaz 2.0 Nov 19 '11 at 3:17

For number 5: I think it's better to distribute the $\cos x$ then observe that the product will result to $1/\sin x$ which is $\csc x$.

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If you do that, you have $\sin\,x+\dfrac{\cos^2 x}{\sin\,x}$, so you still have to do some manipulation instead of merely "observing"... :) –  Ｊ. Ｍ. Nov 19 '11 at 3:29
Yes , it is true but take note that the finding the LCM the numerator will be reduced to 1. so $1/sin x$ –  ken Nov 19 '11 at 4:17