MHB Transforming Trigonometeric Identities II

[Drain Brain]

Transform the left member to the right member.

$\frac{\left(\sin^{2}(\phi)\cos^{2}(\phi)+\cos^{4}(\phi)+2\cos^{2}(\phi)+\sin^{2}(\phi)\right)}{1-\tan^{2}(\phi)}=\frac{3+\tan^{2}(\phi)}{1-\tan^{4}(\phi)}$

I begin by regrouping the numerator of the left member$\frac{\left(\left(\sin^{2}(\phi)\cos^{2}(\phi)+\cos^{4}(\phi)\right)+2\cos^{2}(\phi)+\sin^{2}(\phi)\right)}{1-\tan^{2}(\phi)}$

then factoring out $\cos^{2}(\phi)$ I get,$\frac{\left(\cos^{2}(\phi)\left(\sin^{2}(\phi)+\cos^{2}(\phi)\right)+2\cos^{2}(\phi)+\sin^{2}(\phi)\right)}{1-\tan^{2}(\phi)}$

using Pythagorean identity

$\frac{\left(\cos^{2}(\phi)(1)+2\cos^{2}(\phi)+\sin^{2}(\phi)\right)}{1-\tan^{2}(\phi)}=\frac{\left(3\cos^{2}(\phi)+\sin^{2}(\phi)\right)}{1-\tan^{2}(\phi)}$knowing that $\cot^{2}(\phi)\sin^{2}(\phi)=\cos^{2}(\phi)$ and $\tan^{2}(\phi)(\cos^{2}(\phi)=\sin^{2}(\phi)$ I will replace the terms in the numerator with these relations.

$\frac{\left(3\cot^{2}(\phi)\sin^{2}(\phi)+\tan^{2}(\phi)(\cos^{2}\right)}{1-\tan^{2}(\phi)}$

I will now express $\cot^{2}(\phi)$ in terms of tan,$\frac{\left(\frac{3}{tan^{2}(\phi)}\sin^{2}(\phi)+\tan^{2}(\phi)\cos^{2}(\phi)\right)}{1-\tan^{2}(\phi)}$

factoring out $\frac{3}{tan^{2}(\phi)}$

$\frac{\frac{1}{tan^{2}(\phi)}\left(3\sin^{2}(\phi)+\tan^{4}(\phi)\cos^{2}(\phi)\right)}{1-\tan^{2}(\phi)}$

placing $\tan^{2}(\phi)$ in the numerator and distributing into the denominator I have

$\frac{\left(3\sin^{2}(\phi)+\tan^{4}(\phi)\cos^{2}(\phi)\right)}{\tan^{2}(\phi)-\tan^{4}(\phi)}$

further manipulation of the second term in the numerator and factoring out $\sin^{2}(\phi)$ I wound up with,

$\frac{\sin^{2}(\phi)\left(3+\tan^{2}(\phi)\right)}{\tan^{2}(\phi)-\tan^{4}(\phi)}$

From here I couldn't go any further. What should I do?

 

[MarkFL]

You are good up to this point:

$$\frac{\sin^2(\phi)\left(3+\tan^{2}(\phi)\right)}{\tan^{2}(\phi)-\tan^{4}(\phi)}$$

Divide numerator and denominator by $\sin^2(\phi)$ to get:

$$\frac{3+\tan^{2}(\phi)}{\sec^2(\phi)\left(1-\tan^{2}(\phi)\right)}$$

Now, apply a Pythagorean identity to $\sec^2(\phi)$, and what is the result?

 

[soroban]

Hello, Drain Brain!

Prove: $\;\dfrac{\sin^2\!x\cos^2\!x+\cos^4\!x+2\cos^2\!x+\sin^2\!x}{1-\tan^2\!x)}\:=\:\dfrac{3+\tan^2\!x}{1-\tan^4\!x}$

You did fine, up to: $\;\dfrac{3\cos^2\!x+\sin^2\!x}{1-\tan^2\!x}$

Multiply by $\frac{1+\tan^2\!x}{1+\tan^2\!x}$

$\quad\dfrac{1+\tan^2\!x}{1+\tan^2\!x}\cdot\dfrac{3\cos^2\!x + \sin^2\!x}{1-\tan^2\!x} \;=\;\dfrac{\sec^2\!x(3\cos^2\!x + \sin^2\!x)}{1-\tan^4\!x}$

$\quad =\;\dfrac{3\sec^2\!x\cos^2\!x +\sec^2\!x\sin^2\!x}{1-\tan^4\!x} \;=\; \dfrac{3 + \tan^2\!x}{1-\tan^4\!x}$