How did 15 degree come in the following equation.

http://dc474.4shared.com/download/8_...55515-f1b2071a

plz help.

thanks in advance

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- Dec 19th 2011, 08:55 PMSiddSinverse of trigonometry
How did 15 degree come in the following equation.

http://dc474.4shared.com/download/8_...55515-f1b2071a

plz help.

thanks in advance - Dec 19th 2011, 10:41 PMCaptainBlackRe: inverse of trigonometry
- Dec 21st 2011, 06:31 AMsbhatnagarRe: inverse of trigonometry
$\displaystyle \theta = \tan^{-1}{\left( \frac{1-\sqrt{3}}{1+\sqrt{3}}\right)}=\tan^{-1}{\left( \frac{1-\sqrt{3}}{1+(1)(\sqrt{3})}\right)}$

Using the formula: $\displaystyle \tan^{-1}(x)-\tan^{-1}(y)=\tan^{-1}{\left( \frac{x-y}{1+xy}\right)}$

we get $\displaystyle \theta= \tan^{-1}(1)-\tan^{-1}(\sqrt{3})= 45^{\circ}-60^{\circ}=-15^{\circ}$