# If cot θ = 7/8, evaluate:

(i) (1 + sin θ)(1 - sin θ) / (1 + cos θ)(1 - cos θ)

(ii) cot^{2}θ

**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

AC^{2} = AB^{2} + BC^{2}

= (7k)^{2 }+ (8k)^{2}

= 49k^{2} + 64k^{2}

= 113k^{2}

AC = √113k^{2}

= √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 - sin^{2}θ / 1 - cos^{2}θ [Since, (a + b)(a - b) = (a^{2} - b^{2})]

= [1 - (8/√113)^{2}] / [1 - (7/√113)^{2}]

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

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

= 49/64

(ii) cot^{2}θ = (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 cot^{2}θ = 49/64.

**☛ Related Questions:**

- If 3 cot A = 4, check whether (1 - tan2 A) / (1 + tan2 A) = cos2 A - sin2 A or not.
- In the triangle ABC right-angled at B, if tan A = 1/√3 find the value of:(i) sin A cos C + cos A sin C(ii) cos A cos C - sin A sin C
- In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
- State whether the following are true or false. Justify your answer.(i) The value of tan A is always less than 1(ii) sec A= 12/5 for some value of angle A.(iii) cos A is the abbreviation used for the cosecant of an angle(iv) cot A is the product of cot and A(v) sinθ = 4/3 for some angle θ.

visual curriculum