Question

if cosA + sec A = 2 then cos6A + sec6A = ?​

Answer 1

Answer:

2

Step-by-step explanation:

cos A + sec A = 2

Squaring both the sides,

→ (cos A + sec A)² = 2²

→ cos² A + sec² A + 2 sec A cos A = 4

Since, cos A = 1/sec A

Hence,

→ cos² A + sec² A + 2 sec A × 1/sec A = 4

→ cos² A + sec² A + 2 = 4

→ cos² A + sec² A = 2

Squaring both the sides again,

→ (cos² A + sec² A) = 2²

→ (cos² A)² + (sec²A)² + 2 cos²A sec² A = 4

→ cos⁴ A + sec⁴ A + 2 = 4

→ cos⁴ A + sec⁴ A = 2

Now,

cos⁶ A + sec⁶ A = (cos²)³ A + (sec²)³ A

We know,

a³ + b³ = (a+b)(a²+b²-ab)

→ (cos²)³ A + (sec²)³ A = (cos² A + sec² A)[(cos²)² A + (sec²)² A - cos A sec A ]

→ (cos²)³ A + (sec²)³ A = (cos² A + sec² A)[cos⁴ A + sec⁴ A - cos A sec A ]

→ cos⁶ A + sec⁶ A = (2)[2-1 ]

→ cos⁶ A + sec⁶ A = (2)[1]

→ cos⁶ A + sec⁶ A = 2

Hence, the answer is 2.