# How to have idea to prove trigonometric identities

Hi to explain this better I'll take an example. I have this identity that's giving me a hard time.

$$\frac{\cos^2(a)-\sin^2(b)}{\sin^2(a)\sin^2(b)} = \cot^2(a)\cot^2(b)-1$$

This is what i would do

$$\cos^2(a)/(\sin^2(a)·\sin^2(b))-\sin^2(b)/(\sin^2(a)·\sin^2(b)) \\ \cot^2(a)·1/\sin^2(b)-1/\sin^2(a)$$

then, we know that

$$1=\cos^2(b)+\sin^2(b) \\ \vdots \\ \cot^2(a)·\cot^2(b)+\cot^2(a)-1/\sin^2(a)$$

Which of course is wrong but just wanted to show you guys how my mind thinks. Is there any right way of solving this or do I just have to keep trying. THANKS.

• Use MathJax to format your questions. I've edited your first equation to show you how it's done. – user137731 Nov 2 '15 at 17:02
• @Bye_World Oh, sorry Bye_World. Didn’t read that. Would be more educational your way, yes. – k.stm Nov 2 '15 at 17:05
• @k.stm No worries. OP can still convert all of those fractions to something that looks a little nicer using \frac {a}{b}. – user137731 Nov 2 '15 at 17:07

When you get to here

$$\cot^2(a)\cdot \frac{1}{\sin^2(b)}-\frac 1{\sin^2(a)} = \cot^2(a)\csc^2(b)-\csc^2(a)$$

you should take a look at what you're trying to get to.

We need to get something minus one, that tells us we should look at the Pythagorean identity:

$$\cos^2(a)+\sin^2(a) =1 \\ \implies \cot^2(a)+1=\csc^2(a)$$

We get the second equation by dividing the first by $\sin^2(a)$.

Plugging this in for $\csc^2(a)$ we get

$$\cot^2(a)\csc^2(b)-\csc^2(a) = \cot^2(a)\csc^2(b)-\cot^2(a)-1$$

Factor the $\cot^2(a)$ from the first two terms to get

$$\cot^2(a)\csc^2(b)-\cot^2(a)-1 = \cot^2(a)(\csc^2(b)-1)-1$$

Then from that same identity from before we see that $\csc^2(b)-1=\cot^2(b)$ and we're done.

Try using the $-1$ on the right handside of the equation: $$\frac{\cos^{2}(a)-\sin^{2}(b)}{\sin^{2}(a)\sin^{2}(b)}+1=\frac{\cos^{2}(a)-\sin^{2}(b)}{\sin^{2}(a)\sin^{2}(b)}+\frac{\sin^{2}(a)\sin^{2}(b)}{\sin^{2}(a)\sin^{2}(b)}$$ Now, use the fondamental formula $\sin^{2}(a)=1-\cos^{2}(a)$ and you get $$\frac{\cos^{2}(a)-\sin^{2}(b)+(1-\cos^{2}(a))\sin^{2}(b)}{\sin^{2}(a)\sin^{2}(b)}=\frac{\cos^{2}(a)-\sin^{2}(b)+\sin^{2}(b)-\sin^{2}(b)\cos^{2}(a)}{\sin^{2}(a)\sin^{2}(b)}$$ hence: $$\frac{\cos^{2}(a)-\sin^{2}(b)+\sin^{2}(b)-\sin^{2}(b)\cos^{2}(a)}{\sin^{2}(a)\sin^{2}(b)}=\frac{\cos^{2}(a)(1-\sin^{2}(b))}{\sin^{2}(a)\sin^{2}(b)}=\cot^{2}(a)\cot^{2}(b)$$

$$\frac{\cos^2(a)-\sin^2(b)}{\sin^2(a)\sin^2(b)} = \cot^2(a)\cot^2(b)-1\Longleftrightarrow$$ $$\cos^2(a)-\sin^2(b) = \sin^2(a)\sin^2(b)\left(\cot^2(a)\cot^2(b)-1\right)\Longleftrightarrow$$ $$\cos^2(a)-\sin^2(b) = \sin^2(a)\sin^2(b)\left(\left(\frac{\cos(a)}{\sin(a)}\right)^2\cdot\left(\frac{\cos(b)}{\sin(b)}\right)^2-1\right)\Longleftrightarrow$$ $$\cos^2(a)-\sin^2(b) = \sin^2(a)\sin^2(b)\left(\frac{\cos^2(a)\cos^2(b)}{\sin^2(a)\sin^2(b)}-1\right)\Longleftrightarrow$$ $$\cos^2(a)-\sin^2(b) = \cos^2(a)\cos^2(b)-\sin^2(a)\sin^2(b)\Longleftrightarrow$$ $$\cos^2(a)-1-\cos^2(b) = \cos^2(a)\cos^2(b)-\sin^2(a)\sin^2(b)\Longleftrightarrow$$ $$\cos^2(a)+\cos^2(b)-1 = \cos^2(a)\cos^2(b)-\sin^2(a)\sin^2(b)\Longleftrightarrow$$ $$-1+\cos^2(a)+\cos^2(b)=-1+\cos^2(a)+\cos^2(b)$$

First use $\cot x=\frac{\cos x}{\sin x}$ and work with the RHS .

$$\cot^2 a \cdot \cot^2 b-1=\frac{\cos^2 a \cdot \cos^2 b}{\sin^2 a \cdot \sin^2 b}-1=\frac{\cos^2a \cdot \cos^2 b-\sin^2 a \cdot \sin^2 b}{\sin^2 a \cdot \sin^2 b}$$

All that remains to prove is that $$\cos^2a \cdot \cos^2 b-\sin^2 a \cdot \sin^2 b=\cos^2 a-\sin^2 b$$ but this follows using the classical $\cos^2x+\sin^2x=1$ as follows : $$\cos^2a \cdot \cos^2 b-\sin^2 a \cdot \sin^2 b=\cos^2 a \cdot (1-\sin^2 b)-(1-\cos^2a) \cdot \sin^2 b=\cos^2 a-\sin^2 b$$ as wanted .