Question

Find acute angle such that 5 tan square theta + 3 = 9 sec theta​

Answer 1

Answer:

Step-by-step explanation:

Answer:

θ = 60°

Step-by-step explanation:

Find the acute angle theta such that 5tan square theta + 3 = 9 sec theta​

5Tan²θ + 3 = 9Secθ

Using

Sec²θ = Tan²θ + 1

=> Tan²θ = Sec²θ - 1

5(Sec²θ - 1) + 3 = 9 Secθ

=> 5Sec²θ - 9Secθ - 2 = 0

=> 5Sec²θ - 10Secθ + Secθ   - 2 = 0

=> 5Secθ(Secθ - 2) + 1(Secθ   - 2) = 0

=>  (5Secθ + 1)(Secθ   - 2) = 0

Secθ = -1/5  => Cosθ = -5  (not possible)

Secθ  = 2 => Cosθ = 1/2  => θ = 60°

Answer 2

we know that,

1 + tan²x = sec²x

so,

tan²x = sec²x - 1

$5 {tan}^{2} x + 3 = 9secx \\ 5( {sec}^{2} x - 1) + 3 = 9secx \\ 5 {sec}^{2} x - 5 + 3 = 9secx \\ 5{sec}^{2} x - 9secx - 2 = 0 \\ 5 {sec}^{2} x - 10secx + secx - 2 = 0 \\ 5secx(secx - 2) + 1(secx - 2) = 0 \\ (5secx + 1)(secx - 2) = 0 \\ \\ 5secx + 1 = 0 \\ \: or \\ secx - 2 = 0 \\ \\ secx = \frac{ - 1}{5} \\ or \\ secx = 2 \\ \\ that \: is \\ \\ cosx = - 5 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: or \: \: \: \: \: \: \: \: \: \: \: \: \: \: cosx = \frac{1}{2}$

but,

-1 ≤ cosx ≤ +1

hence, cosx ≠ -5

hence, cosx = 1/2 = cos60°

hence, x = 60°

Hence, the acute angle is 60°