# Simplish trig identity problems

• Nov 9th 2007, 11:57 AM
jono
Simplish trig identity problems
Prove -
1) $\displaystyle sec (u) + tan (u) = \frac {1} {sec(u) - tan(u)}$

2) $\displaystyle sec (u) + cosec (u)cot(u) = sec(u)cosec^2(u)$

3) $\displaystyle sin^2 (u)(1+sec^2 (u) ) = sec^2 (u) - cos^2 (u)$

4) $\displaystyle \frac {1-cos (u)} { sin(u)} = \frac {1} {cosec(u) + cot(u)}$

Any help appreciated
• Nov 9th 2007, 01:10 PM
Soroban
Hello, Jono!

Quote:

$\displaystyle 1)\;\;\sec\theta + \tan\theta \:=\: \frac{1}{\sec\theta - \tan\theta}$
Multiply the left side by $\displaystyle \frac{\sec\theta - \tan\theta}{\sec\theta-\tan\theta}$

$\displaystyle (\sec\theta + \tan\theta)\cdot\frac{\sec\theta-\tan\theta}{\sec\theta-\tan\theta}\;\;=\;\;\frac{\overbrace{\sec^2\!\thet a - \tan^2\!\theta}^{\text{This is 1}}}{\sec\theta - \tan\theta} \;\;=\;\;\frac{1}{\sec\theta - \tan\theta}$

Quote:

$\displaystyle 2)\;\;\sec\theta + \csc\theta\cot\theta \:= \:\sec\theta\csc^2\!\theta$
The left side is: .$\displaystyle \frac{1}{\cos\theta} + \frac{1}{\sin\theta}\cdot\frac{\cos\theta}{\sin\th eta} \;\;=\;\;\frac{1}{\cos\theta} + \frac{\cos\theta}{\sin^2\!\theta}$

Get a common denominator: .$\displaystyle \frac{1}{\cos\theta}\cdot{\color{blue}\frac{\sin^2 \!\theta}{\sin^2\!\theta}} + \frac{\cos\theta}{\sin^2\!\theta} \cdot {\color{blue}\frac{\cos\theta}{\cos\theta}}$

. . $\displaystyle = \;\;\frac{\overbrace{\sin^2\!\theta + \cos^2\!\theta}^{\text{This is 1}}}{\sin^2\!\cos\theta} \;=\;\frac{1}{\sin^2\!\theta\cos\theta} \;\;=\;\;\frac{1}{\cos\theta}\cdot\frac{1}{\sin^2\ !\theta} \;\;=\;\;\sec\theta\csc^2\!\theta$

Quote:

$\displaystyle 3)\;\;\sin^2\!\theta(1 + \sec^2\!\theta) \:= \:\sec^2\!\theta - \cos^2\!\theta$
$\displaystyle \text{The right side is:}$ . $\displaystyle \frac{1}{\cos^2\!\theta} - \cos^2\!\theta \;=\; \frac{1-\cos^4\!\theta}{\cos^2\!\theta} \;=\;\frac{\overbrace{(1-\cos^2\!\theta)}^{\text{This is }\sin^2\!\theta}(1 + \cos^2\!\theta)}{\cos^2\!\theta} \;\;=\;\;\frac{\sin^2\!\theta(1 + \cos^2\!\theta)}{\cos^2\!\theta}$

. . $\displaystyle =\;\;\sin^2\!\theta\cdot\frac{\cos^2\!\theta + 1}{\cos^2\!\theta} \;=\;\sin^2\theta\left(\frac{\cos^2\theta}{\cos^2\ theta} + \frac{1}{\cos^2\theta}\right) \;\;=\;\;\sin^2\!\theta(1 + \sec^2\!\theta)$

Quote:

$\displaystyle 4)\;\;\frac{1 -\cos\theta}{\sin\theta} \:= \:\frac{1}{\csc\theta + \cot\theta}$
The right side is: .$\displaystyle \dfrac{1}{\dfrac{1}{\sin\theta} + \dfrac{\cos\theta}{\sin\theta}}$

$\displaystyle \text{Multiply by }\frac{\sin\theta}{\sin\theta}\!:\quad\frac{\sin\t heta(1)}{\sin\theta \left(\dfrac{1}{\sin\theta} + \dfrac{\cos\theta}{\sin\theta}\right) } \;=\; \frac{\sin\theta}{1 + \cos\theta}$

$\displaystyle \text{Multiply by }\frac{1-\cos\theta}{1-\cos\theta}\!:\quad\frac{\sin\theta}{1+\cos\theta} \cdot\frac{1-\cos\theta}{1-\cos\theta} \;=\;\frac{\sin\theta(1 -\cos\theta)}{\underbrace{1-\cos^2\!\theta}_{\text{This is }\sin^2\!\theta}}$

. . . . $\displaystyle = \;\frac{\sin\theta(1-\cos\theta)}{\sin^2\!\theta} \;=\;\frac{1-\cos\theta}{\sin\theta}$