Question

how can i do it please help me tl solve it

Answer 1

$\large \tt AHOY!! \:$

given :- tan ∅ = 1/√5

consider any right angle ∆ABC.

we know that tan ∅ = perpendicular/base

therefore perpendicular = AB = 1 and base = BC = √5

by Pythagoras we get,

=> AC² = AB² + BC²

=> AC² = 1² + √5²

=> AC² = 1 + 5

=> AC² = 6

=> AC = √6

therefore hypotenuse is √6

now we have to find the value of cosec∅ and sec∅

cosec∅ = hypotenuse/perpendicular = √6/1

sec∅ = hypotenuse/base = √6/√5

$\therefore \sf \frac{ { csc}^{2} \theta - {sec}^{2} \theta}{ {csc}^{2} \theta + {sec}^{2} \theta } = \frac{ { (\sqrt{6} )}^{2} - { (\frac{\sqrt{6} }{ \sqrt{5} } )}^{2} }{ { {( \sqrt{6} )}^{2} + ( \frac{ \sqrt{6} }{ \sqrt{5} } )}^{2} } \\ \\ \sf \Large= \frac{6 - \frac{6}{5} }{6 + \frac{6}{5} } \\ \\ = \sf \Large \frac{ \frac{30}{5} - \frac{6}{5} }{ \frac{30}{5} + \frac{6}{5} } \\ \\ \sf \Large = \frac{ \frac{24}{5} }{ \frac{36}{5} } \\ \\ \sf = \frac{24}{ \cancel5} \times \frac{ \cancel5}{36} \\ \\ \sf = \frac{24}{36} = \frac{2}{3}$

$\large \tt HOPE \: THIS \: HELPS!!$