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

Solution:

We will use the concept of trigonometric ratios to solve the given problem.

Let ΔABC be a right-angled triangle such that tan A = 1/√3

tan A = side opposite to ∠A / side adjacent to ∠A = BC/AB = 1/√3

Let BC = k and AB = √3 k, where k is a positive integer.

By applying Pythagoras theorem in ΔABC, we have

AC² = AB² + BC²

= (√3 k)² + (k)²

= 3k² + k²

= 4k²

AC = 2k

Therefore, sin A = side opposite to ∠A / hypotenuse = BC/AC = 1/2

cos A = side adjacent to ∠A / hypotenuse = AB/AC = √3/2

sin C = side opposite to ∠C / hypotenuse = AB/AC = √3/2

cos C = side adjacent to ∠C / hypotenuse = BC/AC = 1/2

(i) sin A cos C + cos A sin C

By substituting the values of the trigonometric functions in the above equation we get,

sin A cos C + cos A sin C = (1/2)(1/2) + (√3/2)(√3/2)

= 1/4 + 3/4

= (1 + 3)/4

= 4/4

= 1

(ii) cos A cos C - sin A sin C

By substituting the values of the trigonometric functions in the above equation we get,

cos A cos C - sin A sin C = (√3/2)(1/2) - (1/2)(√3/2)

= √3/4 - √3/4

= 0

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Video Solution:

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

Maths NCERT Solutions Class 10 Chapter 8 Exercise 8.1 Question 9

Summary:

In the triangle ABC right-angled at B, if tan A = 1/√3, then the value of sin A cos C + cos A sin C = 1, and cos A cos C - sin A sin C = 0 respectively.

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