# Math Help - Identities

1. ## Identities

I have recieved some great help with proving identities but I have a few more of these Trig Identities I'm having trouble with and I was wondering if anyone could help me prove them ??

$1 + 3sin^2xsec^4x = sec^6x - tan^6x$

$\frac{sec^2x - 6tanx + 7}{sec^2x - 5} = \frac{tanx - 4}{tanx + 2}$

$\frac{1}{4sin^2xcos^2X} - \frac{(1-tan^2x)^2}{4tan^2x} = 1$

$1 + \frac{2}{tanx-cotx} + \frac{1}{sec^2x + csc^2x} = (1 + sinxcosx)^2$

2. $1 + 3\sin^2x \sec^4x = \frac{\cos^4x + 3\sin^2x}{\cos^4x}$

$1 + 3\sin^2x \sec^4x = \frac{(1-\sin^2x)^2 + 3\sin^2x}{\cos^4x}$

$1 + 3\sin^2x \sec^4x = \frac{\sin^4x + \sin^2x + 1}{\cos^4x}$

$1 + 3\sin^2x \sec^4x = \frac{(\sin^4x + \sin^2x + 1)\cos^2x}{\cos^6x}$

$1 + 3\sin^2x \sec^4x = \frac{(\sin^4x + \sin^2x + 1)(1-\sin^2x)}{\cos^6x}$

$1 + 3\sin^2x \sec^4x = \frac{-\sin^6x + 1}{\cos^6x}$

$1 + 3\sin^2x \sec^4x = \sec^6x - \tan^6x$

1) learn and know perfectly the trig identities
2) practice a lot

3. Hello,

For the first one, another method :

Note that $\sec^2x-\tan^2x=\frac{1}{\cos^2x}-\frac{\sin^2x}{\cos^2x}=\frac{1-\sin^2x}{\cos^2x}=1$

So $\tan^2x=\sec^2x-1$

Recall that $(a-1)^3=a^3-3a^2+3a-1$

Hence $\boxed{\tan^6x}=(\tan^2x)^3=(\sec^2x-1)^3=\boxed{\sec^6x-3\sec^4x+3\sec^2x-1}$

And finally,

\begin{aligned}
\sec^6x-\tan^6x &=\sec^6x-\sec^6x+3\sec^4x-3\sec^2x+1 \\
&=1+3\sec^4x-3\sec^2x \\
&=1+3\sec^4x-3\sec^4x\cos^2x \\
&=1+3\sec^4x(1-\cos^2x) \\
&=\boxed{1+3\sec^4x\sin^2x} \end{aligned}

4. Originally Posted by Giggly2
$\frac{1}{{\color{red}4sin^2xcos^2X}} - \frac{(1-tan^2x)^2}{4tan^2x} = 1$
Note that $\tan^2x=\frac{\sin^2x}{\cos^2x}=\frac{{\color{red} \sin^2x\cos^2x}}{\cos^4x} \Rightarrow \boxed{\tan^2x\cos^4x=\sin^2x\cos^2x}$

So a reflex will be to multiply the second fraction by $\frac{\cos^4x}{\cos^4x}$, to get the same denominator :

$\frac{1}{4\sin^2x\cos^2x}-\frac{\cos^4x (1-\tan^2x)^2}{4\sin^2x\cos^2x}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Note that $\sin(2x)=2\cos(x)\sin(x) \Rightarrow 4\cos^2x\sin^2x=\sin^2(2x)$

And note that $\cos^4x(1-\tan^2x)^2=\left(\cos^2x(1-\tan^2x)\right)^2$
And since $\tan^2x=\frac{\sin^2x}{\cos^2x}$, we have $\cos^2x(1-\tan^2x)=\cos^2x-\sin^2x$, which is exactly $\cos(2x)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Hence, we have finally :
\begin{aligned}
\frac{1}{4\sin^2x\cos^2x}-\frac{(1-\tan^2x)^2}{4\tan^2x}
&=\frac{1}{\sin^2(2x)}-\frac{\cos^2(2x)}{\sin^2(2x)} \\
&=\frac{1-\cos^2(2x)}{\sin^2(2x)} \\
&=\boxed{1} \end{aligned}