Results 1 to 4 of 4

Thread: Prove Trig Identity

  1. #1
    Raj
    Raj is offline
    Junior Member
    Joined
    Sep 2007
    Posts
    60

    Prove Trig Identity

    $\displaystyle \frac{sin^2x - tanx}{cos^2x - cotx} = tan^2x$

    I've tried and cannot prove it
    Last edited by Raj; Nov 8th 2008 at 12:21 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hi,

    This equality can be written $\displaystyle \sin^2x-\tan x =\tan^2x\left(\cos^2x-\cot x\right)$. You should try to use $\displaystyle \begin{cases}\tan x = \frac{\sin x }{\cos x}\\ \cot x =\frac{1}{\tan x}\end{cases}$ to show that $\displaystyle \tan^2x\left(\cos^2x-\cot x\right)$ equals $\displaystyle \sin^2x-\tan x$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,550
    Thanks
    15
    Awards
    1
    Quote Originally Posted by Raj View Post
    $\displaystyle \frac{sin^2x - tanx}{cos^2x - cotx} = tan^2x$

    I've tried and cannot solve it
    $\displaystyle \frac{sin^2x - tanx}{cos^2x - cotx} = tan^2x$

    Working with Left Hand Side:

    $\displaystyle \frac{\sin^2 x-\frac{\sin x}{\cos x}}{\cos^2 x-\frac{\cos x}{\sin x}}$

    $\displaystyle \frac {\frac{\sin^2 x \cos x - \sin x}{\cos x}}{\frac {\sin x \cos^2 x-\cos x}{\sin x}}$

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

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

    $\displaystyle \frac{\sin^2 x}{\cos^2 x}$

    $\displaystyle \tan^2 x$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Raj
    Raj is offline
    Junior Member
    Joined
    Sep 2007
    Posts
    60
    Thanks masters


    Quote Originally Posted by flyingsquirrel View Post
    Hi,

    This equality can be written $\displaystyle \sin^2x-\tan x =\tan^2x\left(\cos^2x-\cot x\right)$. You should try to use $\displaystyle \begin{cases}\tan x = \frac{\sin x }{\cos x}\\ \cot x =\frac{1}{\tan x}\end{cases}$ to show that $\displaystyle \tan^2x\left(\cos^2x-\cot x\right)$ equals $\displaystyle \sin^2x-\tan x$.
    I was told I can't balance the identity (multiply/divide both sides) because I'm not initially told if it is equal, I have to prove if it is or if it is not.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. prove the trig identity
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jun 9th 2010, 06:33 PM
  2. Prove the following Trig Identity
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: Apr 2nd 2009, 07:56 AM
  3. Prove this trig identity.
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: Nov 26th 2008, 09:38 AM
  4. Prove Trig Identity
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Nov 12th 2008, 11:52 AM
  5. Prove the Trig Identity
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: Feb 24th 2008, 05:22 PM

Search Tags


/mathhelpforum @mathhelpforum