# stuck with 4 questions

• May 12th 2009, 09:55 AM
chaos
stuck with 4 questions
hi there, im currently stuck on 4 questions at the moment

1. 9x^2 -14x + 1 = 0

PROVE THE FOLLOWING IDENTITIES

1. Sin2θSecθSinθTanθ

2. tan2 A= sin2 A / 1 - sin2 A

3. 2 - tan2 A ≡ 2Sec2 A - 3tan2 A

hope someone out there can help would be much appreciated
• May 15th 2009, 05:15 AM
mr fantastic
Quote:

Originally Posted by chaos
hi there, im currently stuck on 4 questions at the moment

1. 9x^2 -14x + 1 = 0

[snip]

Use the quadratic formula. Are you familiar with it?
• May 15th 2009, 07:05 AM
chaos
Quote:

Originally Posted by mr fantastic
Use the quadratic formula. Are you familiar with it?

no im not unfortunatly, can you explain step by step please?
• May 15th 2009, 07:19 AM
Soroban
Hello, chaos!

Quote:

$1)\;\;9x^2 -14x + 1 \:=\: 0$
You don't know the Quadratic Formula ??

$x \;=\;\frac{\text{-}(\text{-}14) \pm\sqrt{(\text{-}14)^2 - 4(9)(1)}}{2(9)} \;=\;\frac{14 \pm\sqrt{160}}{18} \;=\;\frac{14 \pm4\sqrt{10}}{18} \;=\;\frac{7\pm2\sqrt{10}}{9}$

Quote:

Prove the following identity: . $1)\;\;\sin^2\!\theta\sec\theta \:=\:\sin\theta\tan\theta$
The left side is: . $\sin^2\!\theta\sec\theta \:=\:\sin^2\!\theta\cdot\frac{1}{\cos\theta} \;=\;\sin\theta\cdot\frac{\sin\theta}{\cos\theta} \;=\;\sin\theta\tan\theta$

Quote:

$2)\;\;\tan^2\!A\:=\:\frac{\sin^2\!A}{1 - \sin^2\!A}$

$\text{The right side is: }\;\frac{\sin^2\!A}{\underbrace{1-\sin^2\!A}_{\text{This is }\cos^2\!A}} \;=\;\frac{\sin^2\!A}{\cos^2\!A} \;=\;\left(\frac{\sin A}{\cos A}\right)^2 \;=\;\tan^2\!A$

Quote:

$3)\;\;2 - \tan^2\!A \:=\:2\sec^2\!A - 3\tan^2\!A$
We should know that: . $\sec^2\!A \:=\:\tan^2\!A + 1$

The right side is: . $2\sec^2\!A - 3\tan^2\!A \;\;=\;\;2(\tan^2\!A + 1) - 3\tan^2\!A$

. . . . . . . . . . . $= \;\;2\tan^2\!A + 2 - 3\tan^2\!A \;\;=\;\;2 - \tan^2\!A$

• May 15th 2009, 07:23 AM
chaos
thanks alot for the help guys, much appreciated