If cot θ = 7/8, evaluate:
(i) (1 + sin θ)(1 - sin θ) / (1 + cos θ)(1 - cos θ)
(ii) cot2θ
Solution:
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.
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)
= 49/64
(ii) cot2θ = (7/8)2
= 49/64
☛ Check: NCERT Solutions Class 10 Maths Chapter 8
Video Solution:
If cot θ = 7/8, evaluate: (i) (1 + sin θ)(1 - sin θ) / (1 + cos θ)(1- cos θ), (ii) cot²θ
Maths NCERT Solutions Class 10 Chapter 8 Exercise 8.1 Question 7
Summary:
If cot θ = 7/8, then (1 + sin θ)(1 - sin θ) / (1 + cos θ)(1 - cos θ) = 49/64 and cot2θ = 49/64.
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