Question

If cos−1⁡(pa)+cos−1⁡(qb)=α, then p2a2+kcos⁡α+q2b2=sin2⁡α where k is equal to:

• A. (A) −2pqab
• B. (B) 2pqab
• C. (C) −pqab
• D. (D) pqab

Answer 1

Answer: (A)

(A) −2pqab

Answer 2

Explanation:
Given,p2a2+kcos⁡α+q2b2=sin2⁡α......(i)cos−1⁡(pa)+cos−1⁡(qb)=αAs we know,cos−1⁡x+cos−1⁡y=cos−1⁡(xy−1−x2⋅1−y2)cos−1⁡(pqab−1−p2a21−q2b2)=αcos⁡α=(pqab−1−p2a21−q2b2)pqab−cos⁡α=1−p2a21−q2b2Squaring both sides, we get(pqab−cos⁡α)2=(1−p2a21−q2b2)2(pq)2(ab)2+cos2⁡α−2pqabcos⁡α=(1−p2a2)(1−q2b2)(pq)2(ab)2+cos2⁡α−2pqabcos⁡α=1−p2a2−q2b2+(pq)2(ab)2sin2⁡α=p2a2+q2b2−2pqabcos⁡α.....(ii)Comparing equation (i) and (ii), we getk=−2pqabHence, the correct option is (A).