Question

Prove -
cos30°- Sin 20° ÷ cos 40°+ Cos 20°
= 4/√3×cos40°×cos80°​

Answer 1

Answer

( cos30°- Sin 20° ) / ( cos 40°+ Cos 20° )  = 4/√3 × cos40° × cos80°

Given

$\bullet \bf \;\; \dfrac{cos\ 30^0-sin\ 20^0}{cos\ 40^0+cos\ 20^0}=\dfrac{4}{\sqrt{3}} \times cos\ 40^0 \times cos\ 80^0$

To Prove

$\bullet \bf \;\; \dfrac{cos\ 30^0-sin\ 20^0}{cos\ 40^0+cos\ 20^0}=\dfrac{4}{\sqrt{3}} \times cos\ 40^0 \times cos\ 80^0$

Formula Used

$\bullet \bf \;\; sin(90-\theta)=cos\theta\\\\\bullet \bf \;\; cos(90-\theta)=sin \theta$

$\bullet \;\; \bf cos\ A-cos\ B=2.sin\ \bigg(\dfrac{A+B}{2}\bigg).sin\bigg(\dfrac{B-A}{2}\bigg)\\\\\bullet \;\; \bf cos\ A+cos\ B=2.cos\bigg(\dfrac{A+B}{2}\bigg).cos\bigg(\dfrac{A+B}{2}\bigg)$

Solution

$\bf LHS\\\\\implies \rm \dfrac{cos\ 30-sin\ 20}{cos\ 40+cos\ 20}\\\\\implies \rm \dfrac{cos\ 30-sin\ (90-70)}{cos\ 40+cos\ 20}\\\\\implies \rm \dfrac{cos\ 30-cos\ 70}{cos\ 40+cos\ 20}\\\\\implies \rm \dfrac{2\ sin\bigg(\dfrac{30+70}{2}\bigg) . sin \bigg( \dfrac{70-30}{2}\bigg) }{2\ cos\bigg( \dfrac{40+20}{2}\bigg).cos\bigg( \dfrac{40-20}{2}\bigg)}\\\\\implies \rm \dfrac{2\ sin\ 50.sin\ 20}{2\ cos\ 30.cos\ 10}\\\\\implies \rm \dfrac{2\ sin\ 50 .(2\ sin\ 10\ cos\ 10)}{2\ . \dfrac{\sqrt{3}}{2}.cos\ 10}$

$\implies \rm \dfrac{4\ sin\ 50\ sin\ 10}{\sqrt{3}}\\\\\implies \rm \dfrac{\sqrt{3}}{4}\times sin\ 50 \times sin\ 10\\\\\implies \rm \dfrac{\sqrt{3}}{4}\times sin\ (90-40)\times sin\ (90-80)\\\\\implies \bf \dfrac{\sqrt{3}}{4}\times cos\ 40\times cos\ 80\\\\\implies \bf RHS$