if cot thita=1/√3, show that 1-cos^2thita/2-sin^thita =3/5
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Answer:
LHS-
=(sin thita + cos thita)²+(sin thita - cos thita)²/sin² thita - cos² thita
=sin²thita+cos² thita + 2.sin thita.cos thita + sin²thita+cos²thita -2.sin thita. cos thita/sin²thita - cos²thita
=2sin²thita+2cos²thita/sin²thita-cos²thita
=2/cos²thita (sin²thita/cos²thita-1)
=2sec²thita/tan²thita-1=RHS
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