Question

If tanθ =2-√3 , then prove that tan³θ + cot³θ – 2 = 50​

Answer 1

Answer:

Proof is given below...

Step-by-step explanation:

$\tan\theta=2-\sqrt3\\\\L.H.S\\=\tan^3\theta+\cot^3\theta-2\\=\tan^3\theta+\frac{1}{\tan^3\theta}-2\\\\=\frac{\tan^6\theta-2\tan^3\theta+1}{\tan^3\theta}\\\\=\frac{(\tan^3\theta)^2-2(\tan^2\theta)(1)+1^2}{\tan^3\theta}\\\\=\frac{(\tan^3\theta-1)^2}{\tan^3\theta}\\\\=\frac{[(2-\sqrt3)^3-1]^2}{(2-\sqrt3)^3}\\\\=\frac{[8-3\sqrt3-3(4)(\sqrt3)+3(2)(3)-1]^2}{8-3\sqrt3-3(4)(\sqrt3)+3(2)(3)}\\\\=\frac{(8-3\sqrt3-12\sqrt3+18-1)^2}{8-3\sqrt3-12\sqrt3+18}\\\\=\frac{(25-15\sqrt3)^2}{26-15\sqrt3}$

$=\frac{625+225(3)-2(25)(15\sqrt3)}{26-15\sqrt3}\\\\=\frac{625+675-750\sqrt3}{26-15\sqrt3}\\\\=\frac{1300-750\sqrt3}{26-15\sqrt3}\\\\=\frac{50(26-15\sqrt3)}{26-15\sqrt3}=50=R.H.S.$

∴ proved.

Hope it helps!

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