# Prove That: C O T θ Tan ( 90 ° − θ ) − Sec ( 90 ° − θ ) C O S E C θ + √ 3 Tan 12 ° Tan 60 ° Tan 78 ° = 2 - Mathematics

Sum

Prove that:

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

#### Solution

$LHS = \frac{\sec\left( 90° - \theta \right) cosec\theta - \tan\left( 90°- \theta \right) \cot\theta + \cos^2 25° + \cos^2 65°}{3\tan27° \tan63°}$
$= \frac{cosec\theta cosec\theta - \cot\theta \cot\theta + \sin^2 \left( 90° - 25°\right) + \cos^2 65°}{3\tan27°\cot\left( 90° - 63° \right)}$
$= \frac{{cosec}^2 \theta - \cot^2 \theta + \sin^2 65°+ \cos^2 65°}{3\tan27° \cot27°}$
$= \frac{1 + 1}{3 \times \tan27°\times \frac{1}{\tan27°}}$
$= \frac{2}{3}$
= RHS
Concept: Trigonometry
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#### APPEARS IN

RS Aggarwal Secondary School Class 10 Maths
Chapter 7 Trigonometric Ratios of Complementary Angles
Exercise | Q 5.7 | Page 313