# problem with proving identities using basic identities

• Jun 22nd 2010, 11:27 PM
lindros
problem with proving identities using basic identities
i have a problem with this:

prove,
a)$\displaystyle \frac{1-tan^2\theta}{cot^2 \theta-1}=tan^2\theta$

b)$\displaystyle (cos \theta+sin \theta)^2+(cos \theta-sin \theta)^2=2$

any help will appreciate!! thanks.
• Jun 22nd 2010, 11:32 PM
Unknown008
a) $\displaystyle \frac{1-tan^2\theta}{cot^2\theta-1} = \frac{1-\frac{sin^2\theta}{cos^2\theta}}{\frac{cos^2\theta }{sin^2\theta} - 1}$

$\displaystyle = \frac{\frac{cos^2\theta - sin^2\theta}{cos^2\theta}}{\frac{cos^2\theta - sin^2\theta}{sin^2\theta}}$

$\displaystyle =\frac{\frac{1}{cos^2\theta}}{\frac{1}{sin^2\theta }}$

$\displaystyle = \frac{sin^2\theta}{cos^2\theta}$

$\displaystyle =tan^2\theta$
• Jun 22nd 2010, 11:35 PM
Unknown008
b) $\displaystyle (cos\theta + sin\theta)^2 + (cos\theta - sin\theta)^2 = cos^2 \theta + 2cos\theta sin\theta + sin^2\theta + cos^2\theta -2cos\theta sin\theta + sin^2\theta$

$\displaystyle = cos^2 \theta + \cancel{2cos\theta sin\theta} + sin^2\theta + cos^2\theta \cancel{-2cos\theta sin\theta} + sin^2\theta$

$\displaystyle = 2cos^2\theta + 2 sin^2\theta$

$\displaystyle = 2(cos^2\theta + sin^2\theta)$

$\displaystyle = 2(1)$

$\displaystyle = 2$
• Jun 22nd 2010, 11:50 PM
lindros
Quote:

Originally Posted by Unknown008
a) $\displaystyle \frac{1-tan^2\theta}{cot^2\theta-1} = \frac{1-\frac{sin^2\theta}{cos^2\theta}}{\frac{cos^2\theta }{sin^2\theta} - 1}$

$\displaystyle = \frac{\frac{cos^2\theta - sin^2\theta}{cos^2\theta}}{\frac{cos^2\theta - sin^2\theta}{sin^2\theta}}$

$\displaystyle =\frac{\frac{1}{cos^2\theta}}{\frac{1}{sin^2\theta }}$

$\displaystyle = \frac{sin^2\theta}{cos^2\theta}$

$\displaystyle =tan^2\theta$

i want to know,how can $\displaystyle \frac{cos^2\theta-sin^2\theta}{cos^2 \theta}=\frac{1}{cos^2\theta}$ ??
• Jun 22nd 2010, 11:55 PM
Prove It
Because $\displaystyle \frac{\frac{\cos^2{\theta} - \sin^2{\theta}}{\cos^2{\theta}}}{\frac{\cos^2{\the ta} - \sin^2{\theta}}{\sin^2{\theta}}} = \frac{\frac{1}{\cos^2{\theta}}}{\frac{1}{\sin^2{\t heta}}}\left(\frac{\cos^2{\theta} - \sin^2{\theta}}{\cos^2{\theta} - \sin^2{\theta}}\right)$

$\displaystyle = \frac{\frac{1}{\cos^2{\theta}}}{\frac{1}{\sin^2{\t heta}}}$
• Jun 22nd 2010, 11:56 PM
Unknown008
Ok, I'll break up the steps, but this is not what I meant.

$\displaystyle = \frac{\frac{cos^2\theta - sin^2\theta}{cos^2\theta}}{\frac{cos^2\theta - sin^2\theta}{sin^2\theta}}$

$\displaystyle = \frac{cos^2\theta - sin^2\theta}{cos^2\theta} \times \frac{sin^2\theta}{cos^2\theta - sin^2\theta}$

Then:

$\displaystyle = \frac{1}{cos^2\theta} \times \frac{sin^2\theta}{1}$
• Jun 22nd 2010, 11:59 PM
lindros
thanks very much both of you sir!!