# Verify that $\frac{1}{\tan(x) \csc(x)} = \cos(x)$

Please Help! I have been struggling with this problem for far too long. I have tried rewriting $\tan(x)$ as $\frac{\sin(x)}{\cos(x)}$ in the denominator then simplifying the denominator, but I get stuck there.

• What do you get when you simplify the denominator? What is $$\frac{\sin x }{\cos x} \frac{1}{\sin x}$$ – user296602 Mar 7 '16 at 20:08
• That's exactly where I get stuck. – Dora Mar 7 '16 at 20:09
• @Dora what is $\frac{u}{u}$? Now let $u=\sin(x)$. – Bobson Dugnutt Mar 7 '16 at 20:11
• Do you see that the sine terms cancel? – user296602 Mar 7 '16 at 20:11
• my hints: do u know $\tan x=\sin x/\cos x$ & $\cosec x=1/\sin x$ – Bhaskara-III Mar 7 '16 at 21:35

$$\frac{1}{\tan x\csc x}=\frac{1}{\frac{\sin x}{\cos x}\frac1{\sin x}}=\frac{1}{\frac{1}{\cos x}}=\cos x$$
• @Dora When you multiply by something and then divide by the same thing, it is as if you didn't do it at all; therefore we say that the two things cancel. GoodDeeds saw that $\sin(x)$ was both multiplied and divided in the expression, so they cancelled. – Bobson Dugnutt Mar 7 '16 at 20:14
$$\frac{1}{\tan(x)\csc(x)}=\frac{1}{\tan(x)\cdot\frac{1}{\sin(x)}}=\frac{1}{\frac{\tan(x)}{\sin(x)}}=\frac{\sin(x)}{\tan(x)}=\cos(x)$$