Question

If cosec |-sin |=l and sec |- cos |=m, prove that l2m2(l2+m2+3)=1

Answer 1

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1 answer · Mathematics

Best Answer

cosec(A) - sin(A) = l

⇒ 1/sin(A) - sin(A) = l

⇒ l² = 1/sin²(A) + sin²(A) - 2

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sec(A) - cos(A) = m

⇒ 1/cos(A) - cos(A) = m

⇒ m² = 1/cos²(A) + cos²(A) - 2

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l²m² = [1/sin²(A) + sin²(A) - 2] [1/cos²(A) + cos²(A) - 2] =

= 1/(sin²(A)cos²(A)) + cos²(A)/sin²(A) - 2/sin²(A) + sin²(A)/cos²(A) + sin²(A)cos²(A) - 2sin²(A) - 2/cos²(A) - 2cos²(A) + 4 =

= 1/(sin²(A) + 1/cos²(A)) + (1 - sin²(A))/sin²(A) + (1 - cos²(A))/cos²(A) - 2(1/sin²(A) + 1/cos²(A)) - 2(sin²(A) + cos²(A)) + 4 + sin²(A)cos²(A) =

= -1/sin²(A) + 1/cos²(A) + 1/sin²(A) - 1 + 1/cos²(A) - 1 - 2 + 4 + sin²(A)cos²(A) =

= sin²(A)cos²(A)

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l² + m² + 3 = sin²(A) + 1/sin²(A) - 2 + cos²(A) + 1/cos²(A) - 2 + 3 =

= 1 - 4 + 3 + 1/(sin²(A)cos²(A)) =

= 1/(sin²(A)cos²(A))

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⇒ l²m² (l² + m² + 3) = sin²(A)cos²(A) / [sin²(A)cos²(A)] = 1