# Proof and explanation about Trigonometric Identity

I would like to know about the following identities: $$sec^2\theta-tan^2\theta=1; cosec^2\theta-cot^2\theta=1$$ Both identities are not true for all values of $$\theta$$. My point is if the following is true: $$sec^2\theta-tan^2\theta=1; |sec\theta|\geq1$$ $$\forall\theta\in\mathbb{R}-\{(2n+1)\frac{\pi}{2}, n\in\mathbb{I} \}$$ and $$cosec^2\theta-cot^2\theta=1; |cosec\theta|\geq1$$ $$\forall\theta\in\mathbb{R}- \{n\pi, n\in\mathbb{I} \}$$

• For the one with secant & tangent, the excluded values of $\theta$ should be the odd multiples of $\pi/2$, that is, $(2n+1)(\pi/2)$. These are the places where $\cos\theta=0$. Apr 27 '20 at 3:18
• @GerryMyerson Thank you for your reply. But I would like to know if $n$ could be an integer, once, for example, $sin^2x=0$ when $x=\pi n$, for any integer $n$ . Apr 27 '20 at 12:22
• Yes, that's what happens with the one with cosecant and cotangent. Apr 27 '20 at 12:24
• @GerryMyerson Thank you for your answer and sorry for replying to you again. My teacher presented it, but I still don't understand why $n\in \mathbb{I}$ Apr 27 '20 at 12:33
• You start with $\sin^2\theta+\cos^2\theta=1$. Either you can divide by $\sin^2\theta$ to get $\csc^2\theta-\cot^2\theta=1$, or else $\sin^2\theta=0$, which is the same as $\sin\theta=0$, which is the same as $\theta$ is a multiple of $\pi$. Apr 27 '20 at 12:40

Both identities are obtained by dividing the canonical identity $$\cos^2\theta + \sin^2 \theta = 1$$ by either $$\cos^2 \theta$$ or $$\sin^2 \theta$$ on both sides. The reason the expressions you describe are ill-defined is because $$\cos^2 \theta$$ or $$\sin^2 \theta$$ could be zero, exactly at the values you describe.