Verify the Identity $\sec^2(x)-cot^2\left(\frac{\pi}{2}-x\right)=1$

Question

The identity is $$\sec^2 (x)-\cot^2 \left(\frac{\pi}{2}-x\right)=1$$

Now I know I can rewrite $\sec^2(x)$ as $$1+ \tan^2(x)$$

but would it be possible to rewrite $\cot^2$ as $${\frac{1}{\left(\tan^2(\frac{\pi}{2})\right)\left( \tan^2(-x) \right)}}$$ $$over$$ $$1-\tan^2\left(\frac{\pi}{2}\right)\tan^2(x) $$

since tan is just$\frac{1}{\cot}$ or am I going to the wrong direction?

Answer 1

I think you have most difficulty with showing $\cot{(\frac{\pi}{2}-x)} = \tan x$.

If you were to use the definition of $\cot y = \frac 1{\tan y}$ and then apply the angle addition formula for tangent, you get the term $\tan \frac{\pi}{2}$ in the expression. This is undefined. You can get around this by taking limits, but the neater way is to instead apply another definition: $\cot y = \frac{\cos y}{\sin y}$. You should encounter no difficulties in proving that $\cos{(\frac{\pi}{2}-x)} = \sin x$ and that $\sin {(\frac{\pi}{2}-x)} = \cos x$ (here the application of angle addition formula is uncomplicated), so you now write $\cot{(\frac{\pi}{2}-x)} = \frac{\cos{(\frac{\pi}{2}-x)}}{\sin{(\frac{\pi}{2}-x)} } = \frac{\sin x}{\cos x} = \tan x$.

Now all that remains is to show that $\sec^2 x - \tan^2 x = 1$, which is something you can already easily handle.

Answer 2

Hint: $$\cot(\frac{\pi}{2}-x)=\tan x$$

Answer 3

Notice that $cot^2 (\frac{\pi}{2} - x) = tan^2 x$. So using this, $\sec^2 (x)-\cot^2 \left(\frac{\pi}{2}-x\right)=1=sec^2 x-tan^2 x=1 \leftrightarrow tan^2 x +1 = sec^2 x$, which is a well known identity