The problem, math (√2√3)^2/math math=(√2)^2(√3)^22(√2*√3)/math math=232√6/math math=52√6/math Now, math 2√6=√24=4 Ex 81, 9 In triangle ABC, rightangled at B, if tan A = 1/√3, find the value of sin A cos C cos A sin C tan A = 1/√3 (𝑠𝑖𝑑𝑒 sin 60° cos 30° sin 30° cos 60° = √3/2 ×√3/2 (1/2) ×(1/2 ) = 3/41/4 = 4/4 = (ii) 2 tan 2 45° cos 2 30° – sin 2 60 We know that, the values of the trigonometric ratios are sin 60° = √3/2 cos 30° = √3/2 tan 45° = 1 Substitute the values in the given problem 2 tan 2 45° cos 2 30° – sin 2 60 = 2(1) 2 (√3
Simplify 3 3 3 3
