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$$ {1+\sin^{2}\left(x\right) \over \cos^{2}\left(x\right)} = 1 + 2\tan^{2}\left(x\right)$$

This statement is part of a larger problem, but I need to prove that this is true before moving on. I'm assuming that I would first need to prove $1 + \sin^{2}\left(x\right) = 2\sin^{2}\left(x\right) + \cos^{2}\left(x\right)$, but I'm not sure how to prove that. Any help would be greatly appreciated !.

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Use $\sin^2 x + \cos^2 x \equiv 1$. – Daniel Fischer Apr 7 '14 at 18:46

Multiply through by $\cos^2(x)$ to get $$ 1+\sin^2(x)=\cos^2(x)+2\sin^2(x) $$ then use $\sin^2(x)+\cos^2(x)=1$.

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simplest solution I think – Cruncher Apr 7 '14 at 19:39
But this allegedly "simplest" solution presumes one already knows the identity to be proved. Whereas the solution by @lab bhattacharjee is a direct calculation that starts with the left-hand side and arrives at the right-hand side. – murray Apr 7 '14 at 21:20
@murray: I don't understand what you mean by "presumes one already knows the identity to be proved". If we multiply the given equation by $\cos^2(x)$ (which is reversible in all but those cases where neither side is defined), one gets the equation above. Both proofs are essentially the Pythagorean Theorem (as I commented to lab's answer); mine is simply his multiplied by $\cos^2(x)$. I did this to present the more well known $\sin^2(x)+\cos^2(x)=1$ instead of the equally valid, but slightly more arcane, $\tan^2(x)+1=\sec^2(x)$. – robjohn Apr 7 '14 at 22:02
@robjohn In my precalculus class last year, we had to only manipulate one side of the equation because we had to prove that they were equal. Perhaps this is what he is referring to – scrblnrd3 Apr 8 '14 at 2:37
The phrase "presumes one already knows the identity to be proved" means you have already an equality (which may or may not be true and whose proof is being requested); for that the method of multiplying both sides by the same quantity is perfectly legitimate. But it's mathematically more meaningful just to have the expression on one of the two sides -- NOT an equality -- and then to perform operations upon that expression so as to obtain the expression on the other side of the equation. – murray Apr 8 '14 at 14:27





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Both our answers are based on the Pythagorean Theorem, so they are the same in essence. (+1) – robjohn Apr 7 '14 at 20:29

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