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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