plzz do this question ...
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Answers
Answer:
Step-by-step explanation:
α + β = γ
Tan (α+β) = Tan γ
L.H.S:
//Formula: Tan (α+β) = Tan α + Tan β / 1 - Tanα*Tanβ
Tan α = 1/√x³+x²+x) ;
Tan B = √(x/x²+x+1) = √(x²/x³ + x² + x) (note: multiplied numerator and denominator with x)
Tan α + Tan β / 1 - Tanα*Tanβ = (1/√x³+x²+x) + √(x²/x³ + x² + x)
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1 - [(1/√x³+x²+x) * √(x²/x³ + x² + x)]
= (1 + √x²) / (√x³ + x² + x) / 1 - [(√x²)/ (x³+x²+x)]
= (1 + x) / (√x³ + x² + x) / 1 - [x/(x³+x²+x)]
= (1 + x) / (√x³ + x² + x) / [x³+x²+x - x / x³+x²+x]
= (1 + x) / (√x³ + x² + x) / [x³+x² / x³+x²+x]
= (1 + x) / (√x³ + x² + x) / [x²(x+1)/x³+x²+x ]
= (1 + x) / (√x³ + x² + x) * (x³+x²+x) / x²(x+1)
// cut (1+x) on numerator and denominator
= (x³+x²+x) / x² (√x³ + x² + x)
= (√(x³ + x² + x)) ²/ x² (√x³ + x² + x)
= √(x³ + x² + x) / x²
R.H.S:
Tanγ = √ x⁻³ + x⁻² + x⁻¹
= √ [1/x³ + 1/x² + 1/x]
= √ [(1 + x + x²)/x³]
//multiply numerator and denominator with x
= √[x(1 + x + x²) / x³*x]
= √[x³+x²+x/x⁴]
= √(x³ + x² + x) / x²
Hence L.H.S = R.H.S