First slide

Multiple and sub- multiple Angles

Question

$\frac{\sqrt{2} - \sin α - \cos α}{\sin α - \cos α}$ is equal to

Moderate

A

$\sec (\frac{α}{2} - \frac{π}{8})$
B

$\cos (\frac{π}{8} - \frac{α}{2})$
C

$\tan (\frac{α}{2} - \frac{π}{8})$
D

$\cot (\frac{α}{2} - \frac{π}{2})$

Solution

$\frac{\sqrt{2} - \sin α - \cos α}{\sin α - \cos α}$
$= \frac{\sqrt{2} - \sqrt{2} (\frac{1}{\sqrt{2}} \sin α + \frac{1}{\sqrt{2}} \cos α)}{\sqrt{2} (\frac{1}{\sqrt{2}} \sin α - \frac{1}{\sqrt{2}} \cos α)}$
$= \frac{\sqrt{2} - \sqrt{2} \cos (α - \frac{π}{4})}{\sqrt{2} \sin (α - \frac{π}{4})}$
$= \frac{\sqrt{2} (1 - \cos θ)}{\sqrt{2} \sin θ}$
where, $θ = α - \frac{π}{4} = \frac{2 \sin^{2} (\frac{θ}{2})}{2 \sin (\frac{θ}{2}) \cos (\frac{θ}{2})}$
$= \tan (\frac{θ}{2}) = \tan (\frac{α}{2} - \frac{π}{8})$

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