Find the value of cosx if cot x + cosec x = 5 .

\frac{12}{13}

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\frac{5}{13}

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1

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Solution

The correct option is A $\frac{12}{13}$
$c o t x + c o s e c x = 5$
$\frac{c o s x}{s i n x} + \frac{1}{s i n x} = 5$
$\frac{c o s x + 1}{s i n x} = 5$
$1 + c o s x = 5 s i n x$
Squaring both sides
$(1 + c o s x)^{2} = 25 s i n^{2} x$
$1 + c o s^{2} x + 2 c o s x = 25 (1 - c o s^{2} x)$
$1 + c o s^{2} x + 2 c o s x = 25 - 25 c o s^{2} x$
$26 c o s^{2} x + 2 c o s x - 24 = 0$
$13 c o s^{2} x + c o s x - 12 = 0$
$13 c o s^{2} x + 13 c o s x - 12 c o s x - 12 = 0$
$13 c o s x (c o s x + 1) - 12 (c o s x + 1) = 0$
$(13 c o s x - 12) (c o s x + 1) = 0$
$c o s x = \frac{12}{13}, - 1$

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Similar questions

If cot A = ¹²/₅, then the value of (sin A + cos A) x cosec A is :

{cot}^{- 1} 3 + {cosec}^{- 1} \sqrt{5}

If cot A = $\frac{12}{5}$ , then the value of (sin A + cos A) $\times$ cosec A is:

(1 \frac{3}{5} - \frac{2}{3} \div \frac{12}{13} + \frac{7}{5} \times \frac{1}{3})

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