Description : If cosec |-sin |=l and sec |- cos |=m, prove that l2m2(l2+m2+3)=1 -Maths 9th
Last Answer : cosec(A) - sin(A) = l ⇒ 1/sin(A) - sin(A) = l ⇒ l² = 1/sin²(A) + sin²(A) - 2 --------- sec(A) - cos(A) = m ⇒ 1/cos(A) - cos(A) = m ⇒ m² = 1/cos²(A) + cos²(A) - 2 ---------- l²m² = [1/sin²(A) + ... A)) = = 1/(sin²(A)cos²(A)) ------------- ⇒ l²m² (l² + m² + 3) = sin²(A)cos²(A) / [sin²(A)cos²(A)] = 1