Question

Provie that $sinAsin({60}^{\circ }-A)sin({60}^{\circ }+A)=\frac{4}{4}sin3A.$

Hene , deduce that $sin{20}^{\circ }sin{40}^{\circ }sin{60}^{\circ }som{80}^{\circ }=\frac{3}{16}.$

Or

Prove tahat $cotA+cot({60}^{\circ }+A)-cot({60}^{\circ }-A)=3cot3A.$

Answer 1

$LHS=sinAsin({60}^{\circ }-A)sin({60}^{\circ }+A)=\frac{4}{4}sinAsin({60}^{\circ }-A)sin({60}^{\circ }+A)$

$=\frac{1}{4}\left[2sinA\left\{2sin\right({60}^{\circ }-A\left)sin\right({60}^{\circ }+A\left)\right\}\right]$

$=\frac{1}{4}\left[2sinA\left\{cos\right({60}^{\circ }-A-{60}^{\circ }-A)-cos({60}^{\circ }-A+{60}^{\circ }+A\left)\right\}\right]$

$[\because 2sinAsinB=cos(A-B)-cos(a+B\left)\right]$

$=\frac{1}{4}\times 2sinA\left[cos\right(-2A)-cos{120}^{\circ }]$

$=\frac{1}{4}\times 2sinAcos2A)-cos{120}^{\circ }[\because cos(-\theta )=cos\theta \left]\right]$

$=\frac{1}{4}[2sinAcos2A-2sinAcos{120}^{\circ }]$

$=\frac{1}{4}\left[sin\right(A+2A)+sin(A-2A)-2(sinA\left)(-\frac{1}{2})\right]$

$=\frac{1}{4}[sin3A-sinA+sinA]$

$=\frac{1}{4}sin3A=RHSHenceprovide$

Take,$A={20}^{\circ },thenweget$ $sin{20}^{\circ }sin{40}^{\circ }sin{80}^{\circ }=\frac{1}{4}sin{60}^{\circ }=\frac{1}{4}\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{8}$

$sin{20}^{\circ }sin{40}^{\circ }=\frac{\sqrt{3}}{2}sin{80}^{\circ }=\frac{\sqrt{3}}{8}=\frac{\sqrt{3}}{2}$~~~~~~~~~~~[multiplying both sides by$\frac{\sqrt{3}}{2}]$

$\therefore sin{20}^{\circ }sin{60}^{\circ }sin{80}^{\circ }=\frac{3}{16}[\because \frac{\sqrt{3}}{2}sin{60}^{\circ }]$

we have $LHS=cotA+co({60}^{\circ }+A)-cot({60}^{\circ }-A)$

$=\frac{1}{tanA}+\frac{1}{tan({60}^{\circ }+A)}-\frac{1}{tan{60}^{\circ }-A}$

$=\frac{1}{tanA}+\frac{1}{\frac{tan{60}^{\circ }+tanA}{1-tan{60}^{\circ }tanA}}-\frac{1}{\frac{tan{60}^{\circ }-tanA}{1+tan{60}^{\circ }tanA}}$

$=\frac{1}{tanA}+\frac{1-\sqrt{3}tanA}{\sqrt{3}tanA}+-\frac{1+\sqrt{3}tanA}{\sqrt{3}-tanA}$

$[\because tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}andtan{60}^{\circ }=\sqrt{3}]$

$=\frac{1}{tanA}+\frac{(\sqrt{3}-tanA)(1-\sqrt{3}tanA)-(1+\sqrt{3}tan)(\sqrt{3}+tanA)}{(\sqrt{3}+tanA)(\sqrt{3}-tanA)}$

$=\frac{1}{tanA}+\frac{\sqrt{3}-tanA-3tanA+\sqrt{3}ta{n}^{2}A-(\sqrt{3}+tanA+3tanA+\sqrt{3}ta{n}^{2}A)}{(\sqrt{3}{)}^2-(tanA{)}^2}$

$=\frac{1}{tanA}-\frac{8tanA}{3-ta{n}^{2}A}=\frac{3-ta{n}^{2}A-8tanA.tanA}{tanA(3-ta{n}^{2}A)}$

$=\frac{3-9ta{n}^{2}A}{3tanA-ta{n}^{3}A}=\frac{3(1-ta{n}^{2}A)}{3tanA-ta{n}^{3}A}$

$=\frac{3}{tan3A}[\because tan3A=\frac{3tanA-ta{n}^{3}A}{1-3ta{n}^{2}A}]$

$=3cot3A=RHSHenceproved$