Prove that : tan 11π/3 - 2sin 4π/6 - 3/4cosec^2 π /4 + 4cos^2 17π /6 = (3 - 4√3)/2

Question

Prove that :

tan \(\cfrac{11\pi}3\) - 2sin \(\cfrac{4\pi}6\) - \(\cfrac34\)cosec² \(\cfrac{\pi}4\) + 4cos²\(\cfrac{17\pi}6\)= \(\cfrac{3-4\sqrt3}2\)

Answer 1

LHS =  tan \(\cfrac{11\pi}3\) - 2sin \(\cfrac{4\pi}6\) - \(\cfrac34\)cosec² \(\cfrac{\pi}4\) + 4cos² \(\cfrac{17\pi}6\)

= tan (90° × 7 + 30°) – 2 sin (90° × 1 + 30°) – 3/4 [cosec 45°]² + 4 [cos (90° × 5 + 60°)]²

We know that tan and cos is negative at 90° + θ i.e. in Q₂ and when n is odd, tan → cot, sin → cos and cos → sin.

= [-cot 30°] – 2 cos 30° - 3/4 [cosec 45°]² + [-sin 60°]²

= - cot 30° - 2 cos 30° - 3/4 [cosec 45°]² + [sin 60°]²

= RHS

Hence proved.