# prove the following identities.

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• Jun 22nd 2010, 10:43 PM
mastermin346
prove the following identities.
please help me with this question:

a)$\displaystyle \frac{cot\theta}{tan\theta}+1=cosec^2\theta$

b)$\displaystyle \frac{1+sin\theta}{cos\theta}+\frac{cos\theta}{1+s in\theta}=2 sec\theta.$

please help!
• Jun 22nd 2010, 11:03 PM
Prove It
a) $\displaystyle \frac{\cot{\theta}}{\tan{\theta}} + 1 = \frac{\cot{\theta}}{\frac{1}{\cot{\theta}}} + 1$

$\displaystyle = \cot^2{\theta} + 1$

$\displaystyle = \csc^2{\theta}$.
• Jun 22nd 2010, 11:08 PM
Prove It
b) $\displaystyle \frac{1 + \sin{\theta}}{\cos{\theta}} + \frac{\cos{\theta}}{1 + \sin{\theta}} = \frac{\cos{\theta}(1 + \sin{\theta})}{\cos^2{\theta}} + \frac{\cos{\theta}(1 - \sin{\theta})}{(1 + \sin{\theta})(1 - \sin{\theta})}$

$\displaystyle = \frac{\cos{\theta}(1 + \sin{\theta})}{\cos^2{\theta}} + \frac{\cos{\theta}(1 - \sin{\theta})}{1 - \sin^2{\theta}}$

$\displaystyle = \frac{\cos{\theta}(1 + \sin{\theta})}{\cos^2{\theta}} + \frac{\cos{\theta}(1 - \sin{\theta})}{\cos^2{\theta}}$

$\displaystyle = \frac{\cos{\theta}(1 + \sin{\theta}) + \cos{\theta}(1 - \sin{\theta})}{\cos^2{\theta}}$

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

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

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

$\displaystyle = 2\sec{\theta}$.
• Jun 22nd 2010, 11:15 PM
Unknown008
b) Maybe a shorter route...

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

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

$\displaystyle = \frac{2(1+sin\theta)}{cos\theta(1+sin\theta)}$

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

$\displaystyle = 2sec\theta$