Answer :
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.
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