# Prove an identity and solve an equation

• Nov 18th 2009, 08:05 PM
jhunt47
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!
• Nov 19th 2009, 12:19 AM
Hello jhunt47
Quote:

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!

$\displaystyle \tan x + \cot x = \frac{\sin x}{\cos x}+ \frac{\cos x}{\sin x}$
$\displaystyle =\frac{\sin^2x+\cos^2x}{\sin x\cos x}$

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

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

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

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

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

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

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

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

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

• Nov 19th 2009, 07:31 AM
Hello again jhunt47

This is an interesting one!
Quote:

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
$\displaystyle \left(\frac{16}{81}\right)^{1-\sin^2x}=\left(\frac{16}{81}\right)\cdot\left(\fra c{16}{81}\right)^{-\sin^2x}$
So, if we let $\displaystyle y = \left(\frac{16}{81}\right)^{\sin^2x}$, then
$\displaystyle \left(\frac{16}{81}\right)^{1-\sin^2x}=\left(\frac{16}{81}\right)\cdot y^{-1}=\left(\frac{16}{81}\right)\cdot\frac{1}{y}=\fra c{16}{81y}$
and the equation becomes
$\displaystyle y+\frac{16}{81y}=\frac{26}{27}$

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

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

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

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

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

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