# Trig Identities Problem

I am trying to find the trig identity of this problem:

$\sin(x)\cos(x)\tan(x)\sec(x)\csc(x)$

I already know the answer is $\tan(x)$, but everytime I've tried to solve it I always ended up changing it.

• $\sin x$ and $\csc x$ cancel, as do $\cos x$ and $\sec x$. Commented Feb 27, 2014 at 2:53
• sosmath.com/trig/Trig5/trig5/trig5.html
– Zhoe
Commented Feb 27, 2014 at 2:54

$$\sin(x)\cos(x)\tan(x)\sec(x)\csc(x)=\sin(x)\csc(x)\sec(x)\cos(x)\tan(x)\\ =1\cdot1\cdot\tan(x)=\tan(x)$$
This problem is actually pretty easy to solve. $$\sin(x)\cos(x)\tan(x)\csc(x)\sec(x)$$ $$=\sin(x)\cos(x)\tan(x)\times\dfrac{1}{\sin(x)}\times\dfrac{1}{cos(x)}$$ Now you can cancel out the terms $$1\times 1\times \tan(x)$$ $$=tan(x)$$ That is how you get $\sin(x)\cos(x)\tan(x)\csc(x)\sec(x)=\tan(x)$
It's very simple. \require{cancel} \begin{align} \sin x\cos x\tan x\sec x\csc x&=\cancel{\sin x}\cancel{\cos x}\frac{\sin x}{\cos x}\frac{1}{\cancel{\cos x}}\frac{1}{\cancel{\sin x}}\\ &=\frac{\sin x}{\cos x}\\ &=\tan x \end{align}