Question

Prove that:

\[cot\theta \tan\left( 90° - \theta \right) - \sec\left( 90° - \theta \right)cosec\theta + \sqrt{3}\tan12° \tan60° \tan78° = 2\]

Answer 1

\[ LHS = \frac{\sec\left( 90° - \theta \right) cosec\theta - \tan\left( 90°- \theta \right) \cot\theta + \cos^2 25° + \cos^2 65°}{3\tan27° \tan63°}\]

\[ = \frac{cosec\theta cosec\theta - \cot\theta \cot\theta + \sin^2 \left( 90° - 25°\right) + \cos^2 65°}{3\tan27°\cot\left( 90° - 63° \right)}\]

\[ = \frac{{cosec}^2 \theta - \cot^2 \theta + \sin^2 65°+ \cos^2 65°}{3\tan27° \cot27°}\]
\[ = \frac{1 + 1}{3 \times \tan27°\times \frac{1}{\tan27°}}\]
\[ = \frac{2}{3}\]

= RHS