If $ an heta + \cot heta = 21 $, then find the value of $ 6 an^2 heta + 6\cot^2 heta $.

Question

If $	an 	heta + \cot 	heta = 21 $, then find the value of $ 6	an^2 	heta + 6\cot^2 	heta $.

Examoogle · Accepted Answer

The correct answer is Option C: 2634. If x + 1/x = n, then x² + 1/x² = n² - 2. Since cot θ = 1/tan θ, let tan θ = x. Then x + 1/x = 21. We need 6(x² + 1/x²) = 6(21² - 2) = 6(441 - 2) = 6(439) = 2634.

Step-by-Step Solution & Concepts:
Shortcut Trick: If x + 1/x = n, then x² + 1/x² = n² - 2. Since cot θ = 1/tan θ, let tan θ = x. Then x + 1/x = 21. We need 6(x² + 1/x²) = 6(21² - 2) = 6(441 - 2) = 6(439) = 2634.

Formula Used: $$(a+b)^2 = a^2 + b^2 + 2ab$$

Given: tan θ + cot θ = 21

Steps: Identify the given equation: We are given the expression involving trigonometric ratios: $$	an 	heta + \cot 	heta = 21$$. Square both sides: Square both sides of the equation to reach the squared terms: $$(	an 	heta + \cot 	heta)^2 = 21^2$$. Simplify using identity: Since $	an 	heta \cdot \cot 	heta = 1$, the middle term becomes $2(1) = 2$. Thus, $	an^2 	heta + \cot^2 	heta = 441 - 2 = 439$.. Calculate the final value: Now multiply by 6 as requested: $$6(	an^2 	heta + \cot^2 	heta) = 6 	imes 439 = 2634$$

Answer: C is correct because the result of the expression $6(	an^2 	heta + \cot^2 	heta)$ is $6(21^2 - 2) = 6(439) = 2634$.

Exam Tip: This concept of squaring binomial expressions is frequently used in Algebra (e.g., if $x + 1/x = k$, then $x^2 + 1/x^2 = k^2 - 2$) and applies directly to trigonometry whenever reciprocal functions are involved.

Correct Answer

Solve 10,000+ More Questions with AI Tutor

Discussion (0)

Correct Answer

Solve 10,000+ More Questions with AI Tutor

Related Questions

Discussion (0)