LHS = 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 Q 2 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.

prove that :

LHS

= 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.