First slide

Trigonometric Identities

Question

If $x = \sec ϕ - \tan ϕ$ and $y = cosec ϕ + \cot ϕ$ ,then

Moderate

A

$x = \frac{y + 1}{y - 1}$
B

$x = \frac{y - 1}{y + 1}$
C

$y = \frac{1 + x}{1 - x}$
D

$xy + x - y + 1 = 0$

Solution

we have $x = \frac{1 - \sin ϕ}{\cos ϕ}, y = \frac{1 + \cos ϕ}{\sin ϕ}$
Multiplying, we get
$\begin{array}{l} xy = \frac{(1 - \sin ϕ) (1 + \cos ϕ)}{\cos ϕsin ϕ} \\ \Rightarrow xy + 1 = \frac{1 - \sin ϕ + \cos ϕ - \sin ϕcos ϕ + \sin ϕcos ϕ}{\cos ϕsin ϕ} \\ = \frac{1 - \sin ϕ + \cos ϕ}{\cos ϕsin ϕ} \end{array}$
and
$\begin{matrix} x - y = \frac{(1 - \sin ϕ) \sin ϕ - \cos ϕ (1 + \cos ϕ)}{\cos ϕsin ϕ} \\ = \frac{\sin ϕ - \sin^{2} ϕ - \cos ϕ - \cos^{2} ϕ}{\cos ϕsin ϕ} \\ = \frac{\sin ϕ - \cos ϕ - 1}{\cos ϕsin ϕ} = - (xy + 1) \end{matrix}$
Thus, $xy + x - y + 1 = 0, x = \frac{y - 1}{y + 1}, and y = \frac{1 + x}{1 - x}$

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