Math, asked by Jehriknowledge, 1 year ago

plzz do this question ...​

Attachments:

Answers

Answered by spiderman2019
1

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)

                                                       -------------------------------------------

                                                       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

Similar questions