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Sagot :
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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