# Math Help - Prove an identity and solve an equation

1. ## Prove an identity and solve an equation

Prove the following identity:

[tan(x)+cot(x)]/[sec(x)-tan(x)] = sec^2(x) + sec(x)tan(x)csc^2(x)

Solve the Following Expression:

(16/81)^[sin^2(x)] + (16/81)^[1-sin^2(x)] = (26/27)

I need to know HOW to do these as well as the answers, so if you could show your work that would be appreciated. Thanks!

2. Hello jhunt47
Originally Posted by jhunt47
Prove the following identity:

[tan(x)+cot(x)]/[sec(x)-tan(x)] = sec^2(x) + sec(x)tan(x)csc^2(x)

Solve the Following Expression:

(16/81)^[sin^2(x)] + (16/81)^[1-sin^2(x)] = (26/27)

I need to know HOW to do these as well as the answers, so if you could show your work that would be appreciated. Thanks!
$\tan x + \cot x = \frac{\sin x}{\cos x}+ \frac{\cos x}{\sin x}$
$=\frac{\sin^2x+\cos^2x}{\sin x\cos x}$

$=\frac{1}{\sin x\cos x}$
and $\sec x - \tan x = \frac{1}{\cos x}-\frac{\sin x}{\cos x}$
$=\frac{1-\sin x}{\cos x}$
So $\frac{\tan x + \cot x}{\sec x - \tan x}=\frac{1}{\sin x \cos x}\div\frac{1-\sin x}{\cos x}$
$=\frac{1}{\sin x \cos x}\times\frac{\cos x}{1-\sin x}$

$=\frac{1}{\sin x(1-\sin x)}$

$=\frac{1+\sin x}{\sin x(1-\sin x)(1+\sin x)}$

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

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

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

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

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

$=\sec x \tan x \csc^2 x + \sec^2 x$
No time now to look at part 2. I'll do so later if no-one else has done it in the meantime.

3. Hello again jhunt47

This is an interesting one!
Originally Posted by jhunt47
Solve the Following Expression:

(16/81)^[sin^2(x)] + (16/81)^[1-sin^2(x)] = (26/27)

I need to know HOW to do these as well as the answers, so if you could show your work that would be appreciated. Thanks!
First notice that
$\left(\frac{16}{81}\right)^{1-\sin^2x}=\left(\frac{16}{81}\right)\cdot\left(\fra c{16}{81}\right)^{-\sin^2x}$
So, if we let $y = \left(\frac{16}{81}\right)^{\sin^2x}$, then
$\left(\frac{16}{81}\right)^{1-\sin^2x}=\left(\frac{16}{81}\right)\cdot y^{-1}=\left(\frac{16}{81}\right)\cdot\frac{1}{y}=\fra c{16}{81y}$
and the equation becomes
$y+\frac{16}{81y}=\frac{26}{27}$

$\Rightarrow 81y^2+16=78y$

$\Rightarrow 81y^2-78y+16=0$

$\Rightarrow (3y-2)(27y-8)=0$

$\Rightarrow y = \left(\frac{16}{81}\right)^{\sin^2x}= \frac23, \,\frac{8}{27}$
We now note that $\left(\frac{16}{81}\right)^{\frac14}=\frac23$ and $\left(\frac{16}{81}\right)^{\frac34}=\frac{3}{27}$
$\Rightarrow\sin^2x=\frac14,\,\frac34$

$\Rightarrow \sin x = \pm\frac12,\,\pm\frac{\sqrt3}{2}$

$\Rightarrow x = n\pi \pm \frac{\pi}{6}$ or $n\pi \pm \frac{\pi}{3},\;n \in \mathbb{Z}$