Question

if tantheta+cottheta=2 then value of tan^3 theta+1/tan^3 theta

Answer 1

Given,

$\longrightarrow\tan\theta+\cot\theta=2\quad\quad\dots(1)$

Cubing both sides,

$\longrightarrow\left(\tan\theta+\cot\theta\right)^3=2^3$

$\longrightarrow\tan^3\theta+3\tan^2\theta\cot\theta+3\tan\theta\cot^2\theta+\cot^3\theta=8$

Since $\cot\theta=\dfrac{1}{\tan\theta},$

$\longrightarrow\tan^3\theta+3\tan^2\theta\cdot\dfrac{1}{\tan\theta}+3\tan\theta\cdot\dfrac{1}{\tan^2\theta}+\dfrac{1}{\tan^3\theta}=8$

$\longrightarrow\tan^3\theta+3\tan\theta+3\cdot\dfrac{1}{\tan\theta}+\dfrac{1}{\tan^3\theta}=8$

$\longrightarrow\tan^3\theta+3\tan\theta+3\cot\theta+\dfrac{1}{\tan^3\theta}=8$

$\longrightarrow\tan^3\theta+\dfrac{1}{\tan^3\theta}+3\tan\theta+3\cot\theta=8$

$\longrightarrow\tan^3\theta+\dfrac{1}{\tan^3\theta}+3(\tan\theta+\cot\theta)=8$

From (1),

$\longrightarrow\tan^3\theta+\dfrac{1}{\tan^3\theta}+3\times2=8$

$\longrightarrow\tan^3\theta+\dfrac{1}{\tan^3\theta}+6=8$

$\longrightarrow\underline{\underline{\tan^3\theta+\dfrac{1}{\tan^3\theta}=2}}$

Hence 2 is the answer.

Answer 2

Answer:

answer is in the attachement

Step-by-step explanation:

hope it's helpful