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I have a simple trig identity problem that I can't seem to figure out. I keep going off course in identifying the answer. Here's the problem:

$$ \frac {\cos x}{\sec x} + \frac {\sin x}{\csc x} $$

The answer is:

$$ \csc^2x-\cot^2x $$

I just don't see how they got that answer. I got different solutions, but never that one.

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  • $\begingroup$ What are you going from? Because the first expression equates to $1$ the second is obtained by $$\tan^2 x + 1 = \sec^2x$$ divide through by $\tan^2 x$ but this still doesn't answer your question. $\endgroup$ – Chinny84 Oct 27 '15 at 18:34
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$$\frac{\cos x}{\sec x}+\frac{\sin x}{\csc x}=\cos^2 x+\sin^2 x=1$$

and

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

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  • $\begingroup$ You may want to include some sort of argument for your trig functions. $\endgroup$ – JacobCheverie Oct 27 '15 at 18:37
  • $\begingroup$ Yeah, I know this is not properly written :) Edited. $\endgroup$ – I want to make games Oct 27 '15 at 19:28
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We know that $csc(x)=\frac{1}{sin(x)}$ and similarly $sec(x) = \frac{1}{cos(x)}$. Therefore,

$$ \frac{cos(x)}{sec(x)}=\frac{cos(x)}{1/cos(x)}=cos^2(x) $$

Doing the same thing for your other term you get $cos^2(x)+sin^2(x)=1$, which is a basic trig identity. Another basic trig identity is that $csc^2(x)-cot^2(x)=1$. Your problem and your solution both equal 1 and are therefore equivalent.

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HINT: $$\frac{\cos(x)}{\sec(x)}+\frac{\sin(x)}{\csc(x)}=\sin(x)^2+\cos(x)^2$$

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  • $\begingroup$ why did i got $-1$ i don't understand $\endgroup$ – Dr. Sonnhard Graubner Oct 27 '15 at 18:39
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    $\begingroup$ I did not downvote you, but you have a $\csc$ instead of $\cos$ in your answer. $\endgroup$ – Thomas Oct 27 '15 at 19:01
  • $\begingroup$ thank you it was a typo, sorry $\endgroup$ – Dr. Sonnhard Graubner Oct 27 '15 at 19:08
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$$p=\frac{\cos x}{\sec x}+\frac{\sin x}{\csc x}$$ $$p=\frac{\cos x}{\frac1{\cos x}}+\frac{\sin x}{\frac1{\sin x}}$$ Using the algebraic identity $$\frac{a}{(\frac bc)}=\frac{ac}{b}$$ We have $$p=\cos^2x+\sin^2x$$ $$p=1$$ Then we focus on $$q=\csc^2x-\cot^2x$$ $$q=\frac1{\sin^2x}-\frac{\cos^2x}{\sin^2x}$$ $$q=\frac{1-\cos^2x}{\sin^2x}$$ The Pythagorean identity $\sin^2x=1-\cos^2x$ gives $$q=\frac{\sin^2x}{\sin^2x}$$ $$q=1$$ $$p=q$$ QED

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