# Thread: stuck with 4 questions

1. ## 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

2. 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?

3. 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?

4. Hello, chaos!

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

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

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

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

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

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

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

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

5. thanks alot for the help guys, much appreciated