Question

Prove that
[(sin20°)/(√3×cos20°-2×sin40°)]=1

Answer 1

Answer:

See the explanation

Step-by-step explanation:

$\displaystyle \text{LHS} =\frac{\sin20^\circ}{\sqrt 3\cos20^\circ-2\sin 40^\circ}$

       $\displaystyle =\frac{\sin20^\circ}{\sqrt 3\cos20^\circ-2\sin (60^\circ-20^\circ)}$

      $\displaystyle =\frac{\sin20^\circ}{\sqrt 3\cos20^\circ-2(\sin 60^\circ\cos 20^\circ-\sin 20^\circ\cos60^\circ)}$

      $\displaystyle =\frac{\sin20^\circ}{\sqrt 3\cos20^\circ-2(\frac{\sqrt3}{2}\cos 20^\circ-\sin 20^\circ(\frac{1}{2}))}$

      $\displaystyle =\frac{\sin20^\circ}{\sqrt 3\cos20^\circ-\sqrt3\cos 20^\circ+\sin 20^\circ}$

      $=\displaystyle \frac{\sin 20^\circ}{\sin20^\circ}=1=\text{RHS}$

Hence proved