Question

30. (tana + cosecß)2 - (cotß-seca) = 2 tan a cot B (coseca + secß).
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Answer 1

Step-by-step explanation:

We have,

LHS = (tanA + cosec B)2 - (cotB - sec A)2

⇒ LHS = (tan2A + cosec2B + 2tanA cosecB ) - (cot2B + sec2A - 2cotB secA)

⇒ LHS = (tan2A - sec2A)+(cosec2B - cot2B)+2tanA cosecB +2cotB secA

But,  Sec2A - tan2A =1 & cosec2A - cot2 A = 1

∴ LHS = -1 + 1 + 2 tanA cosecB + 2cotB secA

⇒ LHS = 2 (tanA cosecB + cotB secA)

⇒ LHS = 2 tanA cotB(cosecBcotB+secAtanA) [Dividing and multiplying by tanA cotB]

⇒ LHS = 2tan A cotB{1sinBcosBsinB+1cosAsinAcosA} [Since, CosecA.SinA =1 , SecA.cosA =1, (sinA/cosA)= tanA & (cosA/SinA) =cotA ]

⇒ LHS = 2 tanA cotB(1cosB+1sinA) = 2tanA cotB ( secB + cosecA ) = RHS.  Hence, proved.