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


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

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$\sin x$ and $\csc x$ cancel, as do $\cos x$ and $\sec x$. – André Nicolas Feb 27 '14 at 2:53
up vote 0 down vote accepted

$$\sin(x)\cos(x)\tan(x)\sec(x)\csc(x)=\sin(x)\csc(x)\sec(x)\cos(x)\tan(x)\\ =1\cdot1\cdot\tan(x)=\tan(x)$$

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

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