# Math question

• Jul 30th 2012, 05:38 PM
hacker804
Math question
Hi.Can any one help me with these questions.They are really simple but i cannot seem to get them.Thanks
Q1.Show That:
http://i3.lulzimg.com/5b157758d4.gif

Q2.Prove that:
http://i3.lulzimg.com/1192859b6c.gif

Q3.Prove that:
http://i3.lulzimg.com/aca9108a7d.gif

Thanks a lot.
• Jul 30th 2012, 06:51 PM
srirahulan
Re: Math question
Hi,
I think $\displaystyle Tan(270-\theta)=Cot\theta$
L.H.S
=$\displaystyle tan(270°-x)$
=$\displaystyle \frac{sin(270-\theta}{cos(270-\theta)}$
=$\displaystyle \frac{sin270cos\theta-cos270sin\theta}{cos270cos\theat+sin270sin\theta}$
=$\displaystyle \frac{-cos\theta-0}{0-sin\theta}$
=$\displaystyle \frac{cos\theta}{sin\theta}$
=$\displaystyle cot\theta$

$\displaystyle cos(\alpha+\beta)cos(\alpha-\beta)=cos^2\alpha-sin^2\beta=cos^2\beta-sin^2\alpha$

Take L.H.S

=$\displaystyle \frac{cos(\alpha+\beta)cos(\alpha-\beta)}{1}$
multiply by 2

=$\displaystyle \frac{2cos(\alpha+\beta)cos(\alpha-\beta)}{2}$
=$\displaystyle \frac{cos2\alpha+cos2\beta}{2}$
=$\displaystyle \frac{2cos^2\alpha+2Cos^2\beta-2}{2}$
=$\displaystyle cos^2\alpha+cos^2\beta-1$
=$\displaystyle cos^2\alpha-(1-cos^2\beta)$
=$\displaystyle cos^2\alpha-sin^2\beta$
Or
= $\displaystyle cos^2\beta-(1-cos^2\alpha)$
=$\displaystyle cos^2\beta-sin^2\alpha$

I think you have a mistake,I think the answer will be $\displaystyle -cos\theta$
• Jul 30th 2012, 07:24 PM
hacker804
Re: Math question
Thanks a lot for the help.

BTW,can you tell me of a good equation writing software?

Thanks again.
• Jul 30th 2012, 07:49 PM
Prove It
Re: Math question
Quote:

Originally Posted by hacker804
Hi.Can any one help me with these questions.They are really simple but i cannot seem to get them.Thanks
Q1.Show That:
http://i3.lulzimg.com/5b157758d4.gif

Q2.Prove that:
http://i3.lulzimg.com/1192859b6c.gif

Q3.Prove that:
http://i3.lulzimg.com/aca9108a7d.gif

Thanks a lot.

It's easier in Q.1 to do

\displaystyle \displaystyle \begin{align*} \tan{\left(270^{\circ} - \theta\right)} &= \tan{\left(180^{\circ} + 90^{\circ} - \theta\right)} \\ &= \tan{\left(90^{\circ} - \theta\right)} \\ &= \frac{\sin{\left(90^{\circ} - \theta\right)}}{\cos{\left(90^{\circ} - \theta\right)}} \\ &= \frac{\cos{\theta}}{\sin{\theta}} \\ &= \cot{\theta} \end{align*}
• Jul 30th 2012, 10:10 PM
srirahulan
Re: Math question
Good Idea.Nice work.
• Jul 30th 2012, 10:45 PM
hacker804
Re: Math question
Thanks for all the help