In ΔABC right angled at B, if tanA = √3, then . . .

Question : In ΔABC right angled at B, if tanA = √3, then cosA cosC-sinA sinC = 

a) -1

b) 0

c) 1

d) √3/2

Doubt by Pushkar

Solution :

There are two methods to solve this question : 

Method I (Recommended)

tan A = √3 (Given)

tan A = tan 60°
so A = 60°

Now, In ΔABC
∠B = 90° & ∠A = 60°

∠A+∠B+∠C = 180° (ASP)

60°+90°+∠C=180°
150°+∠C=180°
∠C = 180°-150°

∠C = 30°

Now
cosA cosC-sinA sinC
= cos(60°).cos(30°) - sin(60°).sin(30°)
= (1/2).(√3/2)-(√3/2).(1/2)
= √3/4 - √3/4
= 0

Hence, b) would be the correct option. 

Method II

tan A = √3 (Given)
tan A = P/B = BC/AB = √3/1
BC = √3x
AB = x

By Pythagoras Theorem 

AC² = AB²+BC²
AC² = (x)²+(√3x)²
AC²=x²+3x²
AC² = 4x²
AC = 2x

Now
cos A = B/H = AB/AC = x/2x = 1/2

cos C = B/H = BC/AC = √3x/2x = √3/2

Sin A = P/H = BC/AC = √3x/2x = √3/2

Sin C = P/H = AB/AC = x/2x = 1/2

Now
cosA cosC-sinA sinC
= (1/2).(√3/2)-(√3/2).(1/2)
= √3/4 - √3/4
= 0

Hence, b) would be the correct option.