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Machiavelli

If tanθ=sinα ...

Question

If $tan θ = \frac{sin α - cos α}{sin α + cos α}$ then show that $sin α + cos α = \sqrt{2} cos θ$

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Solution

Given:

$tan θ = \frac{sin α - cos α}{sin α + cos α}$
Divide numerator and denominator of $R H S$ by $cos α$
$\Rightarrow tan θ = \frac{tan α - 1}{tan α + 1}$
$\Rightarrow tan θ = \frac{tan α - tan \frac{π}{4}}{1 + tan α tan \frac{π}{4}}$
$\Rightarrow tan θ = tan (α - \frac{π}{4})$
$\Rightarrow θ = (α - \frac{π}{4})$
Apply cosine on both sides
$\Rightarrow cos θ = cos (α - \frac{π}{4})$
$\Rightarrow cos θ = cos α cos \frac{π}{4} + sin α sin \frac{π}{4}$
$(∵ cos (A - B) = cos A cos B + sin A sin B)$
$\Rightarrow cos θ = \frac{1}{\sqrt{2}} cos α + \frac{1}{\sqrt{2}} sin α$
$\Rightarrow \sqrt{2} cos θ = cos α + sin α$

Hence, proved.

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