Question #ba874

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Question #ba874

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Answer 1 · 2017-10-08T15:40:15

$cot a = \sqrt{- \frac{8}{9}}$ $cot a = sqrt (-8/9)$

Explanation:

Given, $\frac{sec a}{tan a} = \frac{1}{3}$ $sec a/tan a = 1/3$

$\Rightarrow \frac{\frac{1}{cos a}}{\frac{sin a}{cos a}} = \frac{1}{3}$ $rArr (1/cos a)/(sin a/cos a) = 1/3$

$\Rightarrow \frac{1}{cos a} . \frac{cos a}{sin a} = \frac{1}{3}$ $rArr 1/cos a. cos a/sin a = 1/3$

$\Rightarrow \frac{1}{sin a} = \frac{1}{3}$ $rArr 1/sin a = 1/3$

$\Rightarrow cos e c a = \frac{1}{3}$ $rArr cosec a = 1/3$ [ squaring both sides]

$\Rightarrow cos e c^{2} a = \frac{1}{9}$ $rArr cosec^2 a = 1/9$

$\Rightarrow 1 + {cot}^{2} a = \frac{1}{9}$ $rArr 1 + cot^2 a = 1/9$ [ $w e k n o w, 1 + {cot}^{2} a = cos e c^{2} a$ $we know, 1 + cot^2 a = cosec^2 a$ ]

$\Rightarrow {cot}^{2} a = \frac{1}{9} - 1$ $rArr cot^2 a = 1/9-1$

$\Rightarrow {cot}^{2} a = - \frac{8}{9}$ $rArr cot^2 a = -8/9$

$\Rightarrow cot a = \sqrt{- \frac{8}{9}}$ $rArr cot a = sqrt (-8/9)$

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Question #ba874

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