Prove that LHS=RHS:(1/sec^2A-cos^2A+1/cosec^2A-sin^2A)sin^2Acos^2A=1-sin^2Acos^2A/2+sin^2Acos^2A?

Plz answer this quest....its urgent!!!

2 Answers

Relevance
  • Siva
    Lv 5
    1 decade ago
    Best Answer

    (1/sec^2A-cos^2A+1/cosec^2A-sin^2A)sin^2Acos^2A=1-sin^2Acos^2A/2+sin^2Acos^2A

    let us take the LHS first

    (1/sec^2A-cos^2A+1/cosec^2A-sin^2A)sin^2Acos^2A

    ({1/sec^2A}-cos^2A+{1/cosec^2A}-sin^2A)sin^2Acos^2A({cos^2A}-cos^2A+{sin^2A}-sin^2A)sin^2Acos^2A

    (cos^2A-cos^2A+sin^2A-sin^2A)sin^2Acos^2A

    (0)sin^2Acos^2A

    0

    now let us take the RHS

    1-sin^2Acos^2A/2+sin^2Acos^2A

    (1-sin^2Acos^2A)/(2+sin^2Acos^2A)

    i'm not sure how to solve the RHS

  • 1 decade ago

    Please Write Complete Quistion's Statement.

Still have questions? Get your answers by asking now.