Math, asked by sharanyalanka7, 2 months ago

sin2θ = k, then the value of tan^3θ/1 + tan^2θ + cot^3θ/1 + cot^2θ =

Answers

Answered by MrImpeccable
22

ANSWER:

Given:

  • sin2θ = k

To Find:

  • Value of tan³θ/1 + tan²θ + cot³θ/1 + cot²θ

Solution:

:\longrightarrow\dfrac{\tan^3\theta}{1+\tan^2\theta}+\dfrac{\cot^3\theta}{1+\cot^2\theta}\\\\\text{We know that, $\it{1+\tan^2\theta=\sec^2\theta \:\:\&\:\:1+\cot^2\theta=\csc^2\theta}$. So,}\\\\:\implies\dfrac{\tan^3\theta}{\sec^2\theta}+\dfrac{\cot^3\theta}{\csc^2\theta}\\\\\text{Also, $\it{\dfrac{1}{\sec\theta}=\cos\theta\:\:\&\:\:\dfrac{1}{\csc\theta}=\sin\theta}$. So,}\\\\:\implies(\tan^3\theta\times\cos^2\theta)+(\cot^3\theta\times\sin^2\theta)

\text{We know that, $\it{\tan\theta=\dfrac{\sin\theta}{\cos\theta}\:\:\&\:\:\cot\theta=\dfrac{\cos\theta}{\sin\theta}}$. So,}\\\\:\implies\left(\dfrac{\sin^3\theta}{\cos^3\theta}\times\cos^2\theta\right)+\left(\dfrac{\cos^3\theta}{\sin^3\theta}\times\sin^2\theta\right)\\\\:\implies\dfrac{\sin^3\theta}{\cos\theta}+\dfrac{\cos^3\theta}{\sin\theta}\\\\\text{Taking LCM}\\\\:\implies\dfrac{\sin^4\theta+\cos^4\theta}{\sin\theta\cos\theta}\\\\\text{Adding and subtracting $2\sin^2\theta\cos^2\theta$ on numerator,}

:\implies\dfrac{(\sin^4\theta+\cos^4\theta+2\sin^2\theta\cos^2\theta)-2\sin^2\theta\cos^2\theta}{\sin\theta\cos\theta}\\\\\text{We know that, $a^2+b^2+2ab=(a+b)^2$. So,}\\\\:\implies\dfrac{(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta}{\sin\theta\cos\theta}\\\\\text{As $\sin^2\theta+\cos^2\theta=1$. So,}\\\\:\implies\dfrac{(1)^2-2\sin^2\theta\cos^2\theta}{\sin\theta\cos\theta} \\\\:\implies\dfrac{1-2\sin^2\theta\cos^2\theta}{\sin\theta\cos\theta}\\\\\text{On multiplying and dividing by 2,}

:\implies\dfrac{2(1-2\sin^2\theta\cos^2\theta)}{2\sin\theta\cos\theta}\\\\:\implies\dfrac{2-4\sin^2\theta\cos^2\theta}{2\sin\theta\cos\theta}\\\\:\implies\dfrac{2-(2\sin\theta\cos\theta)^2}{2\sin\theta\cos\theta}\\\\\text{We know that, $2\sin\theta\cos\theta=\sin2\theta$. So,}\\\\:\implies\dfrac{2-(\sin2\theta)^2}{\sin2\theta}\\\\\text{We are given that $\sin2\theta=k$. So,} \\\\:\implies\dfrac{2-(k)^2}{k}\\\\\bf{:\implies\dfrac{2-k^2}{k}}

Formulae Used:

  • 1 + tan²θ = sec²θ
  • 1 + cot²θ = cosec²θ
  • 1/secθ = cosθ
  • 1/cosecθ = sinθ
  • a² + b² + 2ab = (a + b)²
  • sin²θ + cos²θ = 1
  • 2sinθcosθ = sin2θ
Answered by amansharma264
23

EXPLANATION.

⇒ sin2θ = k.

To find :

⇒ tan³θ/1 + tan²θ + cot³θ/1 + cot²θ.

As we know that,

Formula of :

⇒ 1 + tan²θ = sec²θ.

⇒ 1 + cot²θ = cosec²θ.

Put the value in the equation, we get.

⇒ tan³θ/sec²θ + cot³θ/cosec²θ.

As we know that,

Formula of :

⇒ tanθ = sinθ/cosθ.

⇒ cotθ = cosθ/sinθ.

⇒ secθ = 1/cosθ.

⇒ cosecθ = 1/sinθ.

⇒ sin³θ/cos³θ x cos²θ + cos³θ/sin³θ x sin²θ.

⇒ sin³θ/cosθ + cos³θ/sinθ.

Taking L.C.M in equation, we get.

⇒ sin⁴θ + cos⁴θ/sinθ cosθ.

We can write,

⇒ sin⁴θ + cos⁴θ = [sin²θ + cos²θ]² - 2sin²θ cos²θ.

⇒ sin⁴θ + cos⁴θ = 1 - 2sin²θ cos²θ.

Put the value in the equation, we get.

⇒ 1 - 2sin²θ cos²θ/sinθ cosθ.

⇒ 1/sinθ cosθ - 2sin²θ cos²θ/sinθ cosθ.

⇒ 1/sinθ cosθ - 2sinθ cosθ.

Put the value of sin2θ = k in equation, we get.

⇒ 1/sinθ cosθ - k.

we can multiply and divide 1/sinθ cosθ x 2, we get.

⇒ 2/2sinθ cosθ - k.

⇒ 2/k - k.

⇒ 2 - k²/k.

⇒ tan³θ/1 + tan²θ + cot³θ/1 + cot²θ = 2 - k²/k.

                                                                                                                       

MORE INFORMATION.

Properties of inverse trigonometric functions.

(1) = sin⁻¹x = cosec⁻¹(1/x). = cosec⁻¹(x) = sin⁻¹(1/x).

(2) = cos⁻¹(x) = sec⁻¹(1/x). = sec⁻¹(x) = cos⁻¹(1/x).

(3) = tan⁻¹(x) = cot⁻¹(1/x). = cot⁻¹(x) = tan⁻¹(1/x).

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