Question

1. If 5tanA=4, Find the value of (5sinA-3cosA)/(4cosA+5sinA)
2. Solve for θ, sinθ/(1+cosθ) + (1+cosθ)/sinθ =4, 0°<θ<90°
3. Prove that tan⁡〖θ-cot⁡θ 〗 = (〖2sin〗^2 θ-1)/sinθcosθ
4. Without using trigonometric tables ,show that
tan 10°tan15°tan75°tan80°=1
5. If x=acosθ-bsinθ and y=asinθ + bcosθ prove that x^2+y^2=a^2+b^2

Answer 1

Answer:

1. (5·sin(A) - 3·cos(A)/(4·cos(A) + 5·sin(A))  = 1/8

2. θ = 30°

3. tan(θ) - cot(θ) = (2·sin²(θ) -1)/((cos(θ)×sin(θ))

from tan(θ) - 1/tan(θ) = sin(θ)/cos(θ) - cos(θ)/sin(θ) and sin²(θ) +  cos²(θ) = 1

4. tan10°·tan15°·tan75°·tan80°= 1 from;

sin(α)·sin(β) = 1/2[cos(α - β) - cos(α + β)]

cos(α)·cos(β) = 1/2[cos(α - β) + cos(α + β)]

5. x² + y² = a² + b² where x = a·cosθ - b·sinθ and y = a·sinθ + b·cosθ from;

cos²θ + sin²θ = 1

Step-by-step explanation:

1. Here we have 5·tan(A) = 5·sin(A)/cos(A) = 4

∴ 5·sin(A) = 4·cos(A)

Hence to find the value of (5·sin(A) - 3·cos(A)/(4·cos(A) + 5·sin(A)) we have;

Substituting the value for 5·sin(A) = 4·cos(A) into the above equation in both the numerator and denominator we have;

(4·cos(A) - 3·cos(A)/(4·cos(A) + 4·cos(A)) = cos(A)/(8·cos(A)) = 1/8

Therefore, (5·sin(A) - 3·cos(A)/(4·cos(A) + 5·sin(A)) = 1/8

2. For the equation as follows, we have

$\frac{sin \theta}{1 + cos \theta} + \frac{1 + cos \theta}{sin \theta} = 4$ this gives

$\frac{2sin (\theta/2) cos (\theta/2) }{2 cos^2 (\theta/2)} + \frac{2 cos^2 (\theta/2)}{2sin (\theta/2) cos (\theta/2) } = 4$

$tan\frac{\theta}{2} + \frac{1}{tan\frac{\theta}{2} } = 4$

$tan^2\frac{\theta}{2} + 1 = 4\times tan\frac{\theta}{2}$

$tan^2\frac{\theta}{2} - 4\cdot tan\frac{\theta}{2} + 1 = 0$

We place;

$tan\frac{\theta}{2} = x$

∴ x² - 4·x + 1 = 0

Factorizing we have

(x - (2 - √3))·(x - (2 + √3))

Therefore, tan(θ/2) = (2 - √3) or (2 + √3)

Solving, we have;

θ/2 = tan⁻¹(2 - √3) or tan⁻¹(2 + √3)

Which gives, θ/2 = 15° or 75°

Hence, θ = 30° or 150°

Since 0° < θ < 90°, therefore, θ = 30°

3. We have tan(θ) - cot(θ) = tan(θ) - 1/tan(θ)

Hence, tan(θ) - 1/tan(θ) = sin(θ)/cos(θ) - cos(θ)/sin(θ)

∴  tan(θ) - 1/tan(θ) = (sin²(θ) -  cos²(θ))/(cos(θ)×sin(θ))...........(1)

From sin²(θ) +  cos²(θ) = 1, we have;

cos²(θ) = 1 - sin²(θ), substituting the value of sin²(θ) in the equation (1) above, we have;

(sin²(θ) -  (1 - sin²(θ)))/(cos(θ)×sin(θ)) = (2·sin²(θ) -1)/((cos(θ)×sin(θ))

Therefore;

tan(θ) - cot(θ) = (2·sin²(θ) -1)/((cos(θ)×sin(θ))

4. tan10°·tan15°·tan75°·tan80°= 1

Here we have since;

sin(α)·sin(β) = 1/2[cos(α - β) - cos(α + β)]

cos(α)·cos(β) = 1/2[cos(α - β) + cos(α + β)]

Then;

tan 10°·tan15°·tan75°·tan80° = tan 10°·tan80°·tan15°·tan75°

tan 10°·tan80°·tan15°·tan75° = $\frac{sin(10^{\circ})}{cos(10^{\circ})} \times \frac{sin(80^{\circ})}{cos(80^{\circ})} \times \frac{sin(15^{\circ})}{cos(15^{\circ})} \times \frac{sin(75^{\circ})}{cos(75^{\circ})}$

Which gives;

$\frac{sin(10^{\circ}) \cdot sin(80^{\circ})}{cos(10^{\circ})\cdot cos(80^{\circ})} \times \frac{sin(15^{\circ}) \cdot sin(75^{\circ})}{cos(15^{\circ})\cdot cos(75^{\circ})}$

$=\frac{1/2[cos(80 - 10) - cos(80 + 10)]}{1/2[cos(80 - 10) + cos(80 + 10)]} \times \frac{1/2[cos(75 - 15) - cos(75 + 15)]}{1/2[cos(75 - 15) + cos(75 + 15)]}$

$=\frac{1/2[cos(70) - cos(90)]}{1/2[cos(70) + cos(90)]} \times \frac{1/2[cos(60) - cos(90)]}{1/2[cos(60) + cos(90)]}$

$=\frac{[cos(70)]}{[cos(70) ]} \times \frac{[cos(60)]}{[cos(60) ]} =1$

5. If x = a·cosθ - b·sinθ and y = a·sinθ + b·cosθ

∴ x² + y² = (a·cosθ - b·sinθ)² + (a·sinθ + b·cosθ)²

= a²·cos²θ - 2·a·cosθ·b·sinθ +b²·sin²θ + a²·sin²θ + 2·a·sinθ·b·cosθ + b²·cos²θ

= a²·cos²θ + b²·sin²θ + a²·sin²θ + b²·cos²θ

= a²·cos²θ + b²·cos²θ + b²·sin²θ + a²·sin²θ

= (a² + b²)·cos²θ + (a² + b²)·sin²θ

= (a² + b²)·(cos²θ + sin²θ) since cos²θ + sin²θ = 1, we have

= (a² + b²)×1 = a² + b²