If tan 2 3 θ= − , then prove that tan3
θ + cot3
θ – 2 = 50
Answers
Answer:
option b :- Ø -2=50
Explanation:
because if we prove it it may we give this answer
Explanation:
Answer:
Proof is given below...
Step-by-step explanation:
\begin{gathered}\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}\\\end{gathered}
tanθ=2−
3
L.H.S
=tan
3
θ+cot
3
θ−2
=tan
3
θ+
tan
3
θ
1
−2
=
tan
3
θ
tan
6
θ−2tan
3
θ+1
=
tan
3
θ
(tan
3
θ)
2
−2(tan
2
θ)(1)+1
2
=
tan
3
θ
(tan
3
θ−1)
2
=
(2−
3
)
3
[(2−
3
)
3
−1]
2
=
8−3
3
−3(4)(
3
)+3(2)(3)
[8−3
3
−3(4)(
3
)+3(2)(3)−1]
2
=
8−3
3
−12
3
+18
(8−3
3
−12
3
+18−1)
2
=
26−15
3
(25−15
3
)
2
=
26−15
3
625+225(3)−2(25)(15
3
)
=
26−15
3
625+675−750
3
=
26−15
3
1300−750
3
=
26−15
3
50(26−15
3
)
=50=R.H.S.
∴ proved.
Hope it helps!