2 keagecR) â ((atBLeed) =P st R (aBecAr 20 2)

1.	2 keagecR) â ((atBLeed) =P st R (aBecAr 20 2)
Answer» LHS = (tanA + cosecB)^2 - (cotB-secA)^2 = (tan^2A + 2tanAcosecB + cosec^2B) - (cot^2B - 2cotBsecA + sec^2A) = tan^2A + cosec^2B + 2tanAcosecB - cot^2B - sec^2A + 2cotBsecA Substituting (tan^2A = sec^2A - 1) and (cosec^2B = 1 + cot^2B) in the above step: (sec^2A - 1) + (1 + cot^2B) + 2tanAcosecB - cot^2B - sec^2A + 2cotBsecA = 2tanAcosecB + 2cotBsecA = 2(tanAcosecB + cotBsecA) RHS = 2tanAcotB(cosecA + secB) = 2tanAcotB((1 / sinA) + (1 / cosB)) = 2(tanA / sinA)cotB + 2tanA(cot B / cosB) = 2(sinA / (cosAsinB))cotB + 2tanA(cosB / (sinBcosB)) = 2(1 / cosA)cotB + 2 tanA(1 / sinB) = 2secAcotB + 2tanAcosecB = 2(secAcotB + tanAcosecB)

Answer»

LHS = (tanA + cosecB)^2 - (cotB-secA)^2 = (tan^2A + 2tanAcosecB + cosec^2B) - (cot^2B - 2cotBsecA + sec^2A) = tan^2A + cosec^2B + 2tanAcosecB - cot^2B - sec^2A + 2cotBsecA

Substituting (tan^2A = sec^2A - 1) and (cosec^2B = 1 + cot^2B) in the above step: (sec^2A - 1) + (1 + cot^2B) + 2tanAcosecB - cot^2B - sec^2A + 2cotBsecA = 2tanAcosecB + 2cotBsecA = 2(tanAcosecB + cotBsecA)

RHS = 2tanAcotB(cosecA + secB) = 2tanAcotB((1 / sinA) + (1 / cosB)) = 2(tanA / sinA)cotB + 2tanA(cot B / cosB) = 2(sinA / (cosAsinB))cotB + 2tanA(cosB / (sinBcosB)) = 2(1 / cosA)cotB + 2 tanA(1 / sinB) = 2secAcotB + 2tanAcosecB = 2(secAcotB + tanAcosecB)

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