Question

$\tan(x) = \frac{1}{ \sqrt{7} }$
find the value of
$\frac{ {cosec}^{2}(x) - {sec}^{2} (x) }{ {cosec}^{2}(x) + {sec}^{2} (x) }$

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Answer 1

Step-by-step explanation:

$\blue{\mathfrak{\underline{\large{Given}}}:} \\ tan \: x \: = \frac{1}{ \sqrt{7} } \\ \\ \blue{\mathfrak{\underline{\large{To \: find}}}:} \\ \frac{ {cosec}^{2}x - {sec}^{2} \: x }{{cosec}^{2}x + {sec}^{2} \: x} = ? \\ \\ \blue{\mathfrak{\underline{\large{Finding}}}:} \\ \frac{ {cosec}^{2}x - {sec}^{2} \: x }{{cosec}^{2}x + {sec}^{2} \: x} \\ \blue{\mathfrak{\underline{\large{Divide \: numerator \: and \: denominator}}}} \\ \blue{\mathfrak{\underline{\large{by \: {cosec}^{2}x }}}:}\\ = \frac{ \frac{{cosec}^{2}x - {sec}^{2} \: x}{ {cosec}^{2}x } }{ \frac{{cosec}^{2}x + {sec}^{2} \: x}{ {cosec}^{2}x } } \\\\ = \frac{1 - \frac{ {sec}^{2}x }{ {cosec}^{2}x } }{1 + \frac{ {sec}^{2}x }{ {cosec}^{2}x} } \\\\ = \frac{1 - \frac{ \frac{1}{ {cos}^{2}x } }{ \frac{1}{ { sin }^{2}x } } }{1 + \frac{ \frac{1}{ {cos}^{2}x } }{ \frac{1}{ { sin }^{2}x } }} \\\\ = \frac{1 - \frac{ {sin}^{2}x }{ {cos}^{2}x } }{1 + \frac{ {sin}^{2}x }{ {cos}^{2}x } } \\\\ = \frac{1 - {tan}^{2}x }{1 + {tan}^{2}x} \\\\ = \frac{1 - {( \frac{1}{ \sqrt{7} } )}^{2} }{1 + {( \frac{1}{ \sqrt{7} } )}^{2} } \\ \\ = \frac{1 - \frac{1}{7} }{1 + \frac{1}{7} } \\ \\ = \frac{7 - 1}{7 + 1} \\\\ = \frac{6}{8} \\ \\ = \frac{3}{4} \\ \\ \red{\mathfrak{\underline{\large{Hope \: It \: Helps \: You }}}} \\ \blue{\mathfrak{\underline{\large{Mark \: Me \: Brainliest}}}}$