Question

If tan theta+cot theta=2, Then tan^10 theta+cot^10 theta =???​

Answer 1

Solution :-

Here , in tanθ + cotθ = 2 , cotθ can be written as 1/tanθ .

→ tanθ + cotθ = 2

→ tanθ + 1/tanθ = 2

→ tan²θ + 1/tanθ = 2

→ tan²θ + 1 = 2tanθ

→ tan²θ + 1² - 2tanθ = 0

→ ( tanθ - 1 )² = 0

→ tanθ - 1 = 0

→ tanθ = 1

Now finding tan¹⁰θ + cot¹⁰θ

Substituting cot¹⁰θ = 1/tan¹⁰θ

→ tan¹⁰θ + 1/tan¹⁰θ

Substituting tan θ = 1

→ 1¹⁰ + 1/1¹⁰

→ 1 + 1/1

→ 1 + 1

→ tan¹⁰θ + cot¹⁰θ = 2

Hence , tan¹⁰θ + cot¹⁰θ = 2

Answer 2

Given:

If tan $\[\tan \theta + \cot \theta = 2\]$, Then $\[{\tan ^{10}}\theta + {\cot ^{10}}\theta \]$ theta =?

Solution:

Find the value of θ,

Put, $\[\cot \theta = \frac{1}{{\tan \theta }}\]$ in $\[\tan \theta + \cot \theta = 2\]$

$\[ \Rightarrow \tan \theta + \frac{1}{{\tan \theta }} = 2\]$

$\[ \Rightarrow \frac{{{{\tan }^2}\theta + 1}}{{\tan \theta }} = 2\]$

$\[ \Rightarrow {\tan ^2}\theta + 1 - 2\tan \theta = 0\]$

$\[ \Rightarrow {\left( {\tan \theta - 1} \right)^2} = 0\]$

Solve further,

$\[\begin{array}{l} \Rightarrow \tan \theta = 1\\ \Rightarrow \tan \theta = \tan 45^\circ \\ \Rightarrow \theta = 45^\circ \end{array}\]$

Evaluate $\[{\tan ^{10}}\theta + {\cot ^{10}}\theta \]$ by putting $\[\theta = 45^\circ \]$,

$\[\begin{array}{l} = {\tan ^{10}}45^\circ + {\cot ^{10}}45^\circ \\ = {1^{10}} + {1^{10}}\\ = 1 + 1\\ = 2\end{array}\]$

Hence, the value of $\[{\tan ^{10}}\theta + {\cot ^{10}}\theta \]$ is 2.