Q. 2 D

# prove that :<

LHS = tan (-225o) cot (-405o) – tan (-765o) cot (675o)

We know that tan (-x) = -tan (x) and cot (-x) = -cot (x).

= [-tan (225°)] [-cot (405°)] – [-tan (765°)] cot (675°)

= tan (225°) cot (405°) + tan (765°) cot (675°)

= tan (90° × 2 + 45°) cot (90° × 4 + 45°) + tan (90° × 8 + 45°) cot (90° × 7 + 45°)

= tan 45° cot 45° + tan 45° [-tan 45°]

= 1 × 1 + 1 × (-1)

= 1 – 1

= 0

= RHS

Hence proved.

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