Math, asked by amphererocks, 10 months ago

if tan theta+cot theta=3, find the value of((tan theta)^2+(cot theta)^2) ((tan theta)^3+(cot theta)^3)

Answers

Answered by brainlyaryan12

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$Tan \theta + Cot \theta = 3$

On squaring both sides :

=> $(Tan \theta+Cot \theta)^2 = 3^2$

=> $\small{Tan^2 \theta + Cot^2 \theta + 2Tan \theta×Cot \theta = 3^2}$

=> $Tan^2 \theta + Cot^2 \theta +2 = 9$

=> $\pink{Tan^2 \theta + Cot^2 \theta = 7 }$ ------(1)

On cubing both sides :

=> $(Tan \theta+Cot \theta)^3 = 3^3$

=> $\tiny{Tan^3 \theta+Cot^3 \theta+3Tan \theta×Cot \theta(Tan \theta+Cot \theta) = 27}$

=> $Tan^3 \theta+Cot^3 \theta+3×1(3) = 27$

=> $Tan^3 \theta+Cot^3 \theta = 27-9$

=> $\pink{Tan^3 \theta+Cot^3 \theta = 18}$ ------(2)

According to question:

$\large(Tan^2 \theta+Cot^2 \theta)(Tan^3 \theta+Cot^3 \theta)$

Using (1) and (2)

=> $\huge{(7)(18)}$

=> $\huge\orange{\fbox{\pink{126}}}$

$\huge{\purple{\bigstar{\blue{\text{Hope it helps...}}}}}$

Answered by akansha804

Answer:

The correct answer to this question is:

(tan²θ + cot²θ) * (tan³θ + cot³θ) = 126

Step-by-step explanation:

Given,

tan θ + cot θ = 3 (equation 1)

To find (tan²θ + cot²θ) * (tan³θ + cot³θ) =?

By the squaring of equation 1 on both sides we get,

(tan θ + cot θ )² = 3²

tan²θ + cot²θ + 2 * tan θ cot θ = 9

tan²θ + cot²θ + 2 * 1 = 9

tan²θ + cot²θ = 9 - 2

tan²θ + cot²θ = 7

By cubing equation 1 on both sides we get,

(tan θ + cot θ )³ = 3³

tan³θ + cot³θ + 3 * tan²θ * cot θ + 3 * tan θ * cot²θ = 27

tan³θ + cot³θ + 3 * tan θ * (tan θ * cot θ)+ 3 * (tan θ * cot θ) * cot θ = 27

tan³θ + cot³θ + 3 * tan θ * 1 + 3 * 1 * cot θ = 27

tan³θ + cot³θ + 3 * tan θ + 3 * cot θ = 27

tan³θ + cot³θ + 3 * (tan θ + cot θ) = 27

tan³θ + cot³θ + 3 * 3 = 27

tan³θ + cot³θ + 9 = 27

tan³θ + cot³θ = 27 - 9

tan³θ + cot³θ = 18

∴ (tan²θ + cot²θ) * (tan³θ + cot³θ) = 7 * 18

(tan²θ + cot²θ) * (tan³θ + cot³θ) = 126

Hence, (tan²θ + cot²θ) * (tan³θ + cot³θ) = 126

Click here for more about trigonometry:

https://brainly.in/question/786386

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