. Prove that
sin^2A. tan A + cos²A.cot A + 2 sin A.cos A = tan A + cot A
Answers
Answer:
here is the answer
Sin²A.TanA + Cos²A.CotA + 2SinA.CosA = TanA + CotA
Step-by-step explanation:
Sin²A.TanA + Cos²A.CotA + 2SinA.CosA = TanA + CotA
LHS
= Sin²A.TanA + Cos²A.CotA + 2SinA.CosA
using TanA = SinA/CosA & CotA = CosA/SinA
= Sin²A.SinA/CosA + Cos²A.CosA/SinA + 2SinA.CosA
= Sin³A/CosA + Cos³A./SinA + 2SinA.CosA
= ( Sin⁴A + Cos⁴A + 2Sin²A.Cos²A )/CosASinA
= ( (Sin²A)² + (Cos²A)² + 2Sin²A.Cos²A )/CosASinA
using a² + b² + 2ab = (a + b)²
a = Sin²A & b = Cos²A
= ((Sin²A) + (Cos²A))²/CosA.SinA
using Sin²A + Cos²A = 1
= 1/CosA.SinA
= (Sin²A + Cos²A)/CosASinA
= Sin²A/CosASinA + Cos²A/CosASinA
= SinA/CosA + CosA/SinA
= TanA + CotA
= RHS
QED
Proved
Answer:
L.H.S. = R.H.S.
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