1. prove [tex]\frac{cos^3(\beta )+sin^3(\beta )}{sin(\theta)+cos(\theta)} = 1-sin(\theta)cos(\theta)[/tex]
2. prove [tex](csc(\theta)sec(\theta))^2 -\frac{(1-tan^2(\theta))^2}{tan^2(\theta)} =4[/tex]

Question

15-08-2022
Mathematics

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1. prove [tex]\frac{cos^3(\beta )+sin^3(\beta )}{sin(\theta)+cos(\theta)} = 1-sin(\theta)cos(\theta)[/tex]
2. prove [tex](csc(\theta)sec(\theta))^2 -\frac{(1-tan^2(\theta))^2}{tan^2(\theta)} =4[/tex]

Respuesta :

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oladayoademosu oladayoademosu · Answer 1 · 2022-08-18T13:58:36+02:00

1. It is proven that (cos³θ + sin³θ)/(sinθ + cosθ) = 1 - sinθscosθ

2. It is proven that (cosecθsecθ)² - (1 - tan²θ)²/tan²θ = 4

1. How to prove the trigonometric expression?

(cos³θ + sin³θ)/(sinθ + cosθ) = 1 - sinθ + cosθ

We need to show that L.H.S = R.H.S

(cos³θ + sin³θ)/(sinθ + cosθ)

Using the sum of two cubes, we have x³ + y³ = (x + y)(x² - xy + y²)

So, (cos³θ + sin³θ) = (cosθ + sinθ)(cosθ² - cosθsinθ + sinθ²)

= (cosθ + sinθ)(cosθ² + sinθ² - cosθsinθ)

= (cosθ + sinθ)(1 - cosθsinθ)

So, (cos³θ + sin³θ)/(sinθ + cosθ) = (cosθ + sinθ)(1 - cosθsinθ)/(sinθ + cosθ)

= 1 - sinθscosθ

Thus, it is proven that (cos³θ + sin³θ)/(sinθ + cosθ) = 1 - sinθscosθ

2. How to prove the trigonometric expression?

(cosecθsecθ)² - (1 - tan²θ)²/tan²θ = 4

WE need to show that L.H.S = R.H.S

(cosecθsecθ)² - (1 - tan²θ)²/tan²θ = (1/sinθcosθ)² - (1 - 2tan²θ + tan⁴θ)/tan²θ

= (1/sinθcosθ)² - 1/tan²θ + 2tan²θ/tan²θ - tan⁴θ/tan²θ

= (1/sinθcosθ)² - 1/tan²θ + 2 - tan²θ

= [(1 - (sinθcosθ)²/tan²θ + 2(sinθcosθ)² - (sinθcosθ)²tan²θ]/(sinθcosθ)²

= [(1 - (sinθcosθ)²cos²θ/sin²θ + 2(sinθcosθ)² - (sinθcosθ)²sin²θ/cos²θ]/(sinθcosθ)²

= [(1 - cos⁴θ + 2(sinθcosθ)² - sin⁴θ]/(sinθcosθ)²

= [(1 - (cos⁴θ + sin⁴θ) + 2(sinθcosθ)² ]/(sinθcosθ)²

Since (cos⁴θ + sin⁴θ = (cos²θ)² + (sin²θ)²

Using the sum of two squares a² + b² = (a + b)² - 2ab

(cos²θ)² + (sin²θ)² = (cos²θ + sin²θ)² - 2cos²θsin²θ = 1 - 2cos²θsin²θ (Since cos²θ + sin²θ = 1)

So, (cosecθsecθ)² - (1 - tan²θ)²/tan²θ = [(1 - (cos⁴θ + sin⁴θ) + 2(sinθcosθ)² ]/(sinθcosθ)²

= [(1 - (1 - 2cos²θsin²θ) + 2(sinθcosθ)² ]/(sinθcosθ)²

= [(1 - 1 + 2cos²θsin²θ + 2(sinθcosθ)² ]/(sinθcosθ)²

= [0 + 4(sinθcosθ)² ]/(sinθcosθ)²

= 4(sinθcosθ)² ]/(sinθcosθ)²

= 4

So, it is proven that (cosecθsecθ)² - (1 - tan²θ)²/tan²θ = 4

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1. prove [tex]\frac{cos^3(\beta )+sin^3(\beta )}{sin(\theta)+cos(\theta)} = 1-sin(\theta)cos(\theta)[/tex] 2. prove [tex](csc(\theta)sec(\theta))^2 -\frac{(1-tan^2(\theta))^2}{tan^2(\theta)} =4[/tex]

Respuesta :

1. How to prove the trigonometric expression?

2. How to prove the trigonometric expression?

Otras preguntas

1. prove [tex]\frac{cos^3(\beta )+sin^3(\beta )}{sin(\theta)+cos(\theta)} = 1-sin(\theta)cos(\theta)[/tex]
2. prove [tex](csc(\theta)sec(\theta))^2 -\frac{(1-tan^2(\theta))^2}{tan^2(\theta)} =4[/tex]