inverse of trigonometry

**SiddS** · December 19th 2011, 08:55 PM

How did 15 degree come in the following equation.

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

thanks in advance

**CaptainBlack** · December 19th 2011, 10:41 PM

Originally Posted by SiddS

How did 15 degree come in the following equation.

Untitled.png - 4shared.com - photo sharing - download image
plz help.

thanks in advance

1. It is not 15 but -15 degrees.

2. Because this involves $\sqrt{3}$ you need to think in terms of 30 or 60 degrees, so look up the tangent half angle formula.

CB

**sbhatnagar** · December 21st 2011, 06:31 AM

$\theta = \tan^{-1}{\left( \frac{1-\sqrt{3}}{1+\sqrt{3}}\right)}=\tan^{-1}{\left( \frac{1-\sqrt{3}}{1+(1)(\sqrt{3})}\right)}$

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

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

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