If $\frac{\sin 135 ° - \cos 240 °}{\sin 135 ° + \cos 240 °} = a + \sqrt{b}$ , then $b =$

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Detailed Solution

$\frac{\sin 135 ° - \cos 240 °}{\sin 135 ° + \cos 240 °} = a + \sqrt{b}$

$\frac{\sin (90 ° + 45 °) - \cos (270 ° - 30 °)}{\sin (90 ° + 45 °) + \cos (270 ° - 30 °)} = a + \sqrt{b}$

$\frac{\cos 45 ° + \sin 30 °}{\cos 45 ° - \sin 30 °} = a + \sqrt{b}$

$\frac{\frac{1}{\sqrt{2}} + \frac{1}{2}}{\frac{1}{\sqrt{2}} - \frac{1}{2}} = a + \sqrt{b}$

$\frac{\sqrt{2} + 1}{\sqrt{2} - 1} = a + \sqrt{b}$

$\frac{\sqrt{2} + 1}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = a + \sqrt{b}$

$\frac{3 + 2 \sqrt{2}}{2 - 1} = a + \sqrt{b}$

$3 + \sqrt{8} = a + \sqrt{b}$

$a = 3, b = 8$

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