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Question Number 123160 by liberty last updated on 23/Nov/20

$G i v e n \sin ρ + \cos ρ = \frac{4}{3}, w h e r e$ $0 < ρ < \frac{π}{4} . F i n d t h e v a l u e o f \cos ρ - \sin ρ .$
	Answered by benjo_mathlover last updated on 23/Nov/20

	$\Rightarrow {(\sin ρ + \cos ρ)}^{2} = \frac{16}{9}$ $\Rightarrow 1 + 2 \sin ρ \cos ρ = \frac{16}{9}$ $\Rightarrow 2 \sin ρ \cos ρ = \frac{7}{9}$ $s i n c e 0 < ρ < \frac{π}{4} t h e n \cos ρ > \sin ρ$ $s o \cos ρ - \sin ρ > 0$ $c o n s i d e r \cos ρ - \sin ρ = \sqrt{1 - 2 \sin ρ \cos ρ}$ $\cos ρ - \sin ρ = \sqrt{1 - \frac{7}{9}} = \frac{\sqrt{2}}{3} .$
	Answered by Dwaipayan Shikari last updated on 23/Nov/20

	${(s i n p + c o s p)}^{2} + {(c o s p - s i n p)}^{2} = 2$ $(c o s p - s i n p) = \sqrt{2 - \frac{16}{9}} = \frac{\sqrt{2}}{3} (A s 0 < p < \frac{π}{4})$
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