Question

Given that cosec x + cot x = 3, evaluate the following:
(i) cosec x - cot x
(ii) cos x

Answer 1

Given that cosec $x + \cot x = 3$, evaluate the following:

(i) cosec $x - \cot x$

(ii) $\cos x$

Using the definitions of the trigonometric functions

$$cosecx = \frac{1}{sinx}$$ $$cotx = \frac{cosx}{sinx}$$ $$\frac{1}{sinx} + \frac{cosx}{sinx} = 3$$

Multiplying both sides by $sinx$ ($sinx \neq 0$)

$$1 + cosx = 3sinx$$ $$1 + cosx - 3sinx = 0$$

Using the double-angle formulae

$$sinx = 2sin\frac{x}{2}cos\frac{x}{2}$$ $$cosx = 2cos^2\frac{x}{2} - 1 \quad (cosx + 1 = 2cos^2\frac{x}{2})$$ $$2cos^2\frac{x}{2} - 6sin\frac{x}{2}cos\frac{x}{2} = 0$$

Solving the equation $cos\frac{x}{2}\left(cos\frac{x}{2} - 3\sin\frac{x}{2}\right) = 0$ we get

1) $cos\frac{x}{2} = 0$ (is not a solution because $sinx = 2\sin\frac{x}{2}cos\frac{x}{2} \neq 0$)

2) $cos\frac{x}{2} - 3\sin\frac{x}{2} = 0$

Dividing by $cos\frac{x}{2}$ we get

$$1 - 3tg\frac{x}{2} = 0, \text{ so } tg\frac{x}{2} = \frac{1}{3}$$

Using the double-angle formulae

$$sinx = 2sin\frac{x}{2}cos\frac{x}{2}$$ $$cosx = 2cos^2\frac{x}{2} - 1 = 1 - 2sin^2\frac{x}{2}$$ $$cos^2\frac{x}{2} = \frac{1 + cosx}{2}$$ $$sin^2\frac{x}{2} = \frac{1 - cosx}{2}$$

i.

$$cosecx - cotx = \frac{1}{2sin\frac{x}{2}cos\frac{x}{2}} - \frac{1 - 2sin^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}} = \frac{2sin^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \tan\frac{x}{2} = \frac{1}{3}$$

ii.

$$\tan\frac{x}{2} = \tan\frac{x}{2}$$

Hence

$$cosx = \frac{1 - \tan^2 x}{1 + \tan^2 x} = \frac{1 - \frac{1}{9}}{1 + \frac{1}{9}} = \frac{4}{5}$$

Answer: i. cosecx - cotx = $\frac{1}{3}$ ii. cosx = $\frac{4}{5}$