Proving the identity $(\tan^2(x)+1)(\cos^2(-x)-1)=-\tan^2(x)$

Question

Proving the trigonometric identity $(\tan{^2x}+1)(\cos{^2(-x)}-1)=-\tan{^2x}$ has been quite the challenge. I have so far attempted using simply the basic trigonometric identities based on the Pythagorean Theorem. I am unsure if these basic identities are unsuitable for the situation or if I am not looking at the right angle to tackle this problem.

Answer 1

Expanding gives $\sin^2x-\tan^2x+\cos^2x-1$. The undesired terms cancel because $\sin^2x+\cos^2x=1$.

Answer 2

Recall:

$$\begin{align} \tan^2(x) - \sec^2(x) &= -1 \\ \sin^2(x) + \cos^2(x) &= 1 \end{align}$$

Thus,

$$\begin{align} \tan^2(x) + 1 &= \sec^2(x)\\ \cos^2(x) - 1 &= -\sin^2(x) \end{align}$$

We also note that $\cos(x)$ is an even function, and thus $\cos(-x) = \cos(x)$. Thus, the formula becomes:

$$(\tan^2(x) + 1)(\cos^2(-x) - 1) = -\sec^2(x)\sin^2(x) = - \frac{\sin^2(x)}{\cos^2(x)} = -\tan^2(x)$$

Answer 3

1. $1+\tan^2x=\sec^2x$
2. $\sin^2x+\cos^2x=1$
3. $\cos x=\cos(-x)$ i.e. $\cos x$ is an even function.
4. $\sec x=1/\cos x$

$$\underbrace{\left(1+\tan^2x\right)}_{=\sec^2x}\underbrace{\left(\cos^2(-x)-1\right)}_{\cos x\text{ is even function}}=-\sec^2x\cdot\sin^2x=-\tan^2x $$

Answer 4

Hint:

Multiply the two members by $\cos^2x$ (certainly nonzero):

$$(\sin^2x+\cos^2x)(\cos^2x-1)=-\sin^2x.$$

Answer 5

We know

• $(\tan x)' = 1 + \tan^2 x = \frac{1}{\cos^2 x}$ and
• $\cos (-x) = \cos x$

Now, it follows immediately \begin{eqnarray*} (\tan^2(x)+1)(\cos^2(-x)-1) & = & (\tan x)'\left(\frac{1}{(\tan x)'}-1\right) \\ & = & 1 - (\tan x)' \\ & = & 1 - (1 + \tan ^2 x) \\ & = & - \tan ^2 x \\ \end{eqnarray*}