1
$\begingroup$

How can I prove the following equation?

\begin{eqnarray} \cot ^2x+\sec ^2x &=& \tan ^2x+\csc ^2x\\ {{1}\over{\tan^2x}}+{{1}\over{\cos^2x}} &=& {{\sin^2x}\over{\cos^2x}}+{{1}\over{\sin^2x}}\\ {{\sin^2x+\cos^4x}\over{\sin^2x\cos^2x}} &=& {{\sin^4x+\cos^2x}\over{\sin^2x\ \cos^2x}}\\ \end{eqnarray}

Then, what can I do...?

Thank you for your attention.

$\endgroup$
3
$\begingroup$

Well if nothing else comes to mind try by hand$$\cot^2 x+\sec^2x=\frac{\cos^2x}{\sin^2x}+\frac 1{\cos^2x}=\frac{\cos^4x+\sin^2x}{\cos^2x\sin^2x}$$

and $$\tan^2x+\csc^2x=\frac {\sin^2x}{\cos^2x}+\frac 1{\sin^2x}=\frac{\sin^4x+\cos^2x}{\cos^2x\sin^2x}$$

And these are equal if $$\cos^4x+\sin^2x=\sin^4x+\cos^2x$$

Now there are various ways to see it. Of course it is easier knowing the standard identities and using them, but they all pretty much boil down to $\sin^2x+\cos^2x=1$, which is in turn another way of writing Pythagoras, and which will definitely help here.

Note also that I've ignored any issue of infinite values or discontinuities. I have been careful not to divide by zero, which is quite easy to do by accident when working with trigonometric sums.

$\endgroup$
4
$\begingroup$

HINT:

We know, $$\csc^2x-\cot^2x=1=\sec^2x-\tan^2x$$

$\endgroup$
2
$\begingroup$

Hint: $1+\cot ^2x = \dfrac{\sin^2x+\cos^2x}{\sin^2x} = \csc ^2x$, $1+\tan ^2x = \dfrac{\cos^2x + \sin^2x}{\cos^2x} = \sec ^2x$.

$\endgroup$
2
$\begingroup$

Do you know, $\cot^2(x) + 1 = \csc^2(x)$ and $\tan^2(x) + 1 = \sec^2(x)$

So, $\cot ^2x+\sec ^2x=\csc^2(x)-1+\tan^2(x) + 1=\tan^2(x)+\csc^2(x)$

Thus, we end up with : $\cot ^2x+\sec ^2x=\tan^2(x)+\csc^2(x)$...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.