# Trig identity $\frac{\cos x}{\sec x} + \frac{\sin x}{\csc x} = \csc^2x - \cot^2x$

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:

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

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

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

• 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. – Chinny84 Oct 27 '15 at 18:34

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

and

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

• You may want to include some sort of argument for your trig functions. – JacobCheverie Oct 27 '15 at 18:37
• Yeah, I know this is not properly written :) Edited. – I want to make games Oct 27 '15 at 19:28

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.

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

• why did i got $-1$ i don't understand – Dr. Sonnhard Graubner Oct 27 '15 at 18:39
• I did not downvote you, but you have a $\csc$ instead of $\cos$ in your answer. – Thomas Oct 27 '15 at 19:01
• thank you it was a typo, sorry – Dr. Sonnhard Graubner Oct 27 '15 at 19:08

$$p=\frac{\cos x}{\sec x}+\frac{\sin x}{\csc x}$$ $$p=\frac{\cos x}{\frac1{\cos x}}+\frac{\sin x}{\frac1{\sin x}}$$ 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