Show $\frac{d}{dx} \tan^3{x}-3 \tan{x}+3x = 3 \tan^4{x}$

Solution 1:

No mistakes. Keep going...

From where you left off: $$\eqalign{ {3(\sin^2 x-\cos^2 x+\cos^4 x)\over\cos^4 x} &={3\bigl(\sin^2 x+\cos^2 x(\cos^2 x-1)\bigr)\over\cos^4 x}\cr &={3\bigl(\sin^2 x+\cos^2 x(-\sin^2 x)\bigr)\over\cos^4 x}\cr &={3\bigl(\sin^2x ( 1-\cos^2 x) \bigr)\over\cos^4 x}\cr &={3 \sin^2x (\sin^2 x)\over\cos^4 x}\cr &={3 \sin^4x \over\cos^4 x}\cr &=3\tan^4 x. } $$

Solution 2:

Your approach is correct, but here's another way to get the same answer. Notice that both the derivative (i.e., $3 \tan^2 x \sec^2 x - 3 \sec^2 x + 3$) and the right hand side ($=3 \tan^4 x$) involve only $\tan^2x$ terms. This is an ideal situation to make use of the identity $$\color{blue}{(\ast)} \quad \sec^2 x - \tan^2 x = 1 .$$ Of course, this is just a variant of the usual Pythagoras theorem: $\sin^2 x + \cos^2 x = 1$.

Here's the complete sequence of steps: $$ \begin{align*} 3 \tan^2 x \sec^2 x - 3 \sec^2 x + 3 &= 3 \ \sec^2 x (\tan^2 x -1) + 3 \\ &\stackrel{\color{blue}{(\ast)}}{=} 3 (\tan^2 x + 1) (\tan^2 x -1) + 3 \\ &= 3 \Big((\tan^2 x)^2 - 1^2 \Big) + 3 \\ &= 3 \tan^4 x - 3 + 3 \\ &= 3 \tan^4 x . \end{align*} $$