Question

trigonometry question​

Answer 1

$\huge \fbox \red{Answer}$

Given,

$\tan(θ) = \frac{1}{ \sqrt{2} }$

By using identity, [ Sec²θ - tan²θ = 1 ]

$\sec(θ) = \sqrt{1 + {\tan}^{2} (θ) }$

$= \sqrt{1 + {( \frac{1}{ \sqrt{2} }) }^{ {2} } }$

$= \sqrt{1 + \frac{1}{2} }$

$= \sqrt{ \frac{3}{2} }$

Also,

$\cot(θ) = \frac{1}{ \tan(θ) }$

$= \frac{1}{ \frac{1}{ \sqrt{2} } }$

$= \sqrt{2}$

By using identity,

$\cosec(θ) = \sqrt{1 + {cot}^{2}θ}$

$= \sqrt{1 + 2}$

$= \sqrt{3}$

Substituting the value, we get

$\frac{ {cosec}^{2}θ- {sec}^{2} θ }{ {cosec}^{2}θ - {cot}^{2}θ}$

$\frac{ { (\sqrt{3}) }^{2} - {( \frac{ \sqrt{3} }{2}) }^{2} }{( { \sqrt{3}) }^{2} + {( \sqrt{2}) }^{2} }$

$\frac{3 - \frac{3}{2} }{3 + 2}$

$\frac{ \frac{3}{2} }{5}$

$\frac{3}{10}$

Hence, value of

$\frac{ {cosec}^{2}θ - {sec}^{2}θ}{ {cosec}^{2}θ + {cot}^{2}θ} = \frac{3}{5}$