If tan - - cot - - 2- then the value of tan3 -1tan3 - is

Nbsp;tan θ+1tan^3 nbsp;θ ⇒ nbsp;tan^3 nbsp;θ+1tan^3 nbsp;θ= ( tan θ +1tan θ ) ( tan^2 θ +1tan^2θ 1 ) =2 ( ( tan θ +1tan θ )^2 3 ) =2(4 3) =2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Intersection

If tan θ + co...

Question

If $t a n θ + c o t θ = 2,$ then the value of $t a n^{3} θ + \frac{1}{t a n^{3} θ}$ is

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 2
$t a n θ + \frac{1}{t a n^{3} θ}$

$\Rightarrow t a n^{3} θ + \frac{1}{t a n^{3} θ} = (t a n θ + \frac{1}{t a n θ}) (t a n^{2} θ + \frac{1}{t a n^{2} θ} - 1)$

$= 2 ({(t a n θ + \frac{1}{t a n θ})}^{2} - 3)$

= $2 (4 - 3)$

=2

Suggest Corrections

Similar questions

Q. If

t a n θ + c o t θ = 2,

then the value of

t a n^{3} θ + \frac{1}{t a n^{3} θ}

Q. The value of

\frac{tan 3 θ}{tan θ} = ?

Q. If

\frac{tan 3 θ - 1}{tan 3 θ + 1} = \sqrt{3}

, then the general value of

θ

\frac{{tan}^{3} θ - 1}{tan θ - 1} = {sec}^{2} θ + tan θ

$Solve the following equations: (i) t a n θ + t a n 2 θ + t a n 3 θ = 0 (i i) t a n θ + t a n 2 θ = t a n 3 θ (i i i) t a n 3 θ + t a n θ = 2 t a n 2 θ$

Join BYJU'S Learning Program

If tanθ+cotθ=2, then the value of tan3θ+1tan3θ is

The correct option is C 2tanθ+1tan3θ ⇒tan3θ+1tan3θ=(tanθ+1tanθ)(tan2θ+1tan2θ−1) =2((tanθ+1tanθ)2−3) =2(4−3) =2

If $t a n θ + c o t θ = 2,$ then the value of $t a n^{3} θ + \frac{1}{t a n^{3} θ}$ is

The correct option is C 2
$t a n θ + \frac{1}{t a n^{3} θ}$

$\Rightarrow t a n^{3} θ + \frac{1}{t a n^{3} θ} = (t a n θ + \frac{1}{t a n θ}) (t a n^{2} θ + \frac{1}{t a n^{2} θ} - 1)$

$= 2 ({(t a n θ + \frac{1}{t a n θ})}^{2} - 3)$

= $2 (4 - 3)$

=2