tan (55° — θ) — cot (35° + θ) = θ
Consider L.H.S.
= tan (55° — θ) — cot (35° + θ)
= tan (90°-(35° +θ)) — cot (35° + θ)
= cot (35° + θ) - cot (35° + θ)
= 0
Hence, proved.