JEE Advanced Model Paper 19 with solutions for online practice

Let Z1 = cos A + i sin A , Z2 = cos 2A + i sin 2A and Z3 = cos 4A + i sin 4A
Then Z1,Z2,Z3 = cos (A + 2A + 4A)
∴ Z1,Z2,Z3 = cos 7A = cos 2π + i sin 2π = 1
→ (cos A + i sin A) (cos 2A + i sin 2A) (cos 4A + i sin 4A) = 1
→ cos A , cos 2A , cos 4A (1 + i tan A) (1 + i tan 2A) = 1
Let C = cos

2π

7

. cos

4π

7

. cos

8π

7

= cos A cos 2A cos 4A and
S = sin

2π

7

. sin

4π

7

. sin

8π

7

. Then
8 C . S = sin

4π

7

. sin

16π

7

But sin

16π

7

= sin (2π+

2π

7

)
∴ (1 + i tan A) ( 1 + i tan 2A) (1 + i tan 4A) = 8.
Simplify ans. – 7