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I am currently doing some Trigonometric Equations, most of which is rather simple, however I came across a little problem. It asked to prove $$\sin^2\theta(\cot^2 \theta +1 ) = 1$$

In itself, I found the answer quite quickly but I had trouble finishing it off. I ended up with $LHS = \cos^2 \theta + \sin^2 \theta$

Now, as the common trigonometric identity would say, this would equal 1. And it does! (if you were to put LHS back into the original equation).

The issue here is that I do not understand how to finish it off. How do I merge this result back into the original equation? Is it ok to be using an identity as proof it is equal? How do i explain that it will be equal. This part is tripping me up pretty bad and I would highly appreciate an explanation of this.

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    As you said LHS of $sin^2\theta(cot^2 \theta +1 ) = 1$ is $cos^2 \theta + sin^2 \theta$ which is equal to $1$. And RHS of $sin^2\theta(cot^2 \theta +1 ) = 1$ is $1$. So LHS=RHS. – 2017-02-17
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    It sounds like you proved the left side was equal to the right side. Wasn't that the objective? – 2017-02-17

2 Answers 2

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write $$\sin(\theta)^2\frac{\cos(\theta)^2}{\sin(\theta)^2}+\sin(\theta)^2=\sin(\theta)^2+\cos(\theta)^2=1$$ if $$\sin(\theta)^2\ne 0$$

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One way of going the other way around-

$RHS= 1= \sin \theta\csc \theta= \sin^2 \theta\csc^2 \theta = \sin^2 \theta(cot^2\theta+1) = LHS$

(Given that $ \theta \ne n\pi $ for any integer $n$ as $ \sin \theta \ne 0$ for this to be true)