Question

Question -:

If tanθ + cotθ = 2 ; Find the value of tan^11θ + cot^11θ...

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Answer 1

Solution:-

Given:-

tanθ + cotθ = 2 

To Find:-

tan^11θ + cot^11θ= ?

Find:-

tanθ + cotθ = 2

=) tanθ + 1/tanθ = 2

=) (tan^2θ + 1)/tanθ = 2

=) tan^2θ + 1 = 2tanθ

=) tan^2θ - 2tanθ +1 = 0

We know that,

(a - b)^2 = a^2 + b^2 - 2ab

Here,

a = tanθ

b = 1

Hence, we got

( tanθ - 1)^2 = 0

=) tanθ = 1

we know that,

tan 45° = 1.

=) tanθ = tan 45°

Cancelling tan from both the sides. we get,

=) θ = 45°

Now,

Taking tan^11θ + cot^11θ

=) tan^(11) 45° + cot^(11) 45°

We know that,

tan 45° = 1

cot 45° = 1.

=) 1^(11) + 1^(11)

=) 1 + 1

=) 2.

Hence,

tan^(11) θ + cot^(11) θ = 2.

Answer 2

Solution :

Given, $\displaystyle\mathrm{tan\theta+cot\theta=2}$

$\displaystyle\implies \mathrm{tan\theta+\frac{1}{tan\theta}=2}$

$\displaystyle\implies \mathrm{\frac{tan^{2}\theta+1}{tan\theta}=2}$

$\displaystyle\implies \mathrm{\frac{2\:tan\theta}{1+tan^{2}\theta}=1}$

$\displaystyle\implies \mathrm{\frac{\frac{2\:sin\theta}{cos\theta}}{1+\frac{sin^{2}\theta}{cos^{2}\theta}}=1}$

$\displaystyle\implies \mathrm{\frac{\frac{2\:sin\theta}{cos\theta}}{\frac{sin^{2}\theta+cos^{2}\theta}{cos^{2}\theta}}=1}$

$\displaystyle\implies \mathrm{2\:sin\theta\:cos\theta=1}$

$\displaystyle\implies \mathrm{sin2\theta=1=sin\frac{\pi}{2}}$

$\displaystyle\implies \mathrm{2\theta=\frac{\pi}{2}}$

$\displaystyle\implies \underline{\mathrm{\theta=\frac{\pi}{4}}}$

Now, $\displaystyle\mathrm{tan^{11}\theta+cot^{11}\theta}$

$\displaystyle\implies \mathrm{tan^{11}\theta+cot^{11}\theta=tan^{11}\frac{\pi}{4}+cot^{11}\frac{\pi}{4}}$

$\displaystyle\implies \mathrm{tan^{11}\theta+cot^{11}\theta=1^{11}+1^{11}}$

$\displaystyle\implies \boxed{\mathrm{tan^{11}\theta+cot^{11}\theta=2}}$