marcoking059
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If cot 0=7/8 evaluate : (i) (1 + sin 8)(1-sin 8) (1 + cos 0) (1 - cos8) (ii) cot¹0​

Sagot :

rafikiedu14

Answer: The value of (1 + sin 8)(1-sin 8) (1 + cos 0) (1 - cos8) =0.2086

and [tex]cot^{1}[/tex]θ=7/8

Step-by-step explanation:

Let base and height of a triangle be 7k and 8k

By applying Pythagoras theorem in Δ ABC, we get

[tex]AC^{2} = AB^{2} +BC^{2}[/tex]

= (7k)2 + (8k)2

= 49k2 + 64k2

= 113k2

AC = √113k2

= √113k

so hypotenuse =[tex]\sqrt{133}k[/tex];

now ;

=(1 + sin θ) (1 - sin θ)(1 + cos θ) (1 - cos θ)[since : (a + b)(a - b) = (a2 - b2)]

=(1 - [tex]sin^{2}[/tex]θ)(1 - [tex]cos^{2}[/tex]θ)

=[1 - (8/√113)2] [1 - (7/√113)2]

= (1 - 64/113) (1 - 49/113)

= (49/113) (64/113)

= 0.2086

hence;

The value of (1 + sin 8)(1-sin 8) (1 + cos 0) (1 - cos8) =0.2086

and [tex]cot^{1}[/tex]θ=7/8

Learn more about trignometry here https://brainly.com/question/28493506

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Answer:

Step-by-step explanation:We will use the basic concepts of trigonometric ratios to solve the problem.

Consider ΔABC as shown below where angle B is a right angle.

If cot θ = 7/8, evaluate: (i) (1 + sin θ)(1 - sin θ) / (1 + cos θ)(1- cos θ), (ii) cot²θ

cot θ = side adjacent to θ / side opposite to θ = AB/BC = 7/8

Let AB = 7k and BC = 8k, where k is a positive integer.

By applying Pythagoras theorem in Δ ABC, we get

AC2 = AB2 + BC2

= (7k)2 + (8k)2

= 49k2 + 64k2

= 113k2

AC = √113k2

= √113k

Therefore, sin θ = side opposite to θ / hypotenuse = BC/AC = 8k/√113k = 8/√113

cos θ = side adjacent to θ / hypotenuse = AB/AC = 7k/√113k = 7/√113

(i) (1 + sin θ) (1 - sin θ) / (1 + cos θ) (1 - cos θ) = 1 - sin2θ / 1 - cos2θ  [Since, (a + b)(a - b) = (a2 - b2)]

= [1 - (8/√113)2] / [1 - (7/√113)2]

= (1 - 64/113) / (1 - 49/113)

= (49/113) / (64/113)

(ii) cotФ≈

= 49/64

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