Question

If cosec θ + cot θ = 5, then evaluate sec θ

Answer 1

cosec θ + cot θ = 5

∴\(\frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta} =5\)

∴ 1 + cos θ = 5.sin θ 

Squaring both the sides, we get 

1 + 2 cos θ + cos² θ = 25 sin² θ 

∴ cos² θ + 2 cos θ + 1 = 25 (1 – cos² θ) 

∴ cos² θ + 2 cos θ + 1 = 25 – 25 cos² θ 

∴ 26 cos² θ + 2 cos θ – 24 = 0 

∴ 26 cos² θ + 26 cos θ – 24 cos θ – 24 = 0 

∴ 26 cos θ (cos θ + 1) – 24 (cos θ + 1) = 0 

∴ (cos θ + 1) (26 cos θ – 24) = 0 

∴ cos θ + 1 = θ or 26 cos θ – 24 = 0 

∴ cos θ = -1 or cos θ = \(\frac{24}{26} = \frac {12}{13}\)

When cos θ = -1, sin θ = 0 

∴ cot θ and cosec x are not defined, 

∴ cos θ ≠ -1 

∴ cos θ = \(\frac {12}{13}\)

∴ sec θ = \(\frac {1}{\cos \theta} = \frac {13}{12}\)

[Note: Answer given in the textbook is -1 or \(\frac {13}{12}\). 

However, as per our calculation it is only \(\frac {13}{12}\).]