Question

given that tan theta equal to 1 by root 5 what is the value of cosec square theta minus sec square theta by cosec square theta + sec square theta​

Answer 1

AnswEr :

$\star\:\sf{Given :\tan(\theta) = \dfrac{1}{\sqrt{5}}}$

$\star\:\sf{To \:Find : \dfrac{\csc^{2} (\theta) - \sec^{2}( \theta)}{\csc^{2} (\theta) + \sec^{2}( \theta)}}$

$\rule{150}{2}$

$\underline{\bigstar\:\textsf{According to the Question Now :}}$

$:\implies\dfrac{ \csc^{2} (\theta) - \sec^{2}( \theta)}{\csc^{2} (\theta) + \sec^{2}( \theta)}\\\\ \qquad \scriptsize{\bf{\dag} \:\csc(\theta) = \dfrac{1}{ \sin(\theta) }\: \tt and \: \sec(\theta)=\dfrac{1}{ \cos(\theta)}} \\\\:\implies\dfrac{ \dfrac{1}{\sin^{2} (\theta)} - \dfrac{1}{ \cos^{2} (\theta)}}{\dfrac{1}{\sin^{2} (\theta)} +\dfrac{1}{ \cos^{2} (\theta)}}\\\\\qquad \scriptsize{ \bf{\dag}\:\texttt{Multiplying each term by \sin^2(\theta) }}\\\\:\implies\dfrac{ \bigg(\dfrac{1}{\sin^{2} (\theta)} \times\sin^{2} (\theta) \bigg) - \bigg(\dfrac{1}{ \cos^{2} (\theta)} \times \sin^{2} (\theta) \bigg)}{\bigg(\dfrac{1}{\sin^{2} (\theta)} \times\sin^{2} (\theta) \bigg) + \bigg(\dfrac{1}{ \cos^{2} (\theta)} \times\sin^{2} (\theta) \bigg)}\\\\\qquad \scriptsize{ \bf{\dag}\: \dfrac{\sin^{2}( \theta) }{ \cos^{2}( \theta)} =\tan^{2}(\theta) }\\\\:\implies\dfrac{1 - \tan^{2}(\theta)}{1 + \tan^{2}(\theta)}\\\\\\\:\implies\sf\dfrac{1 - \bigg(\dfrac{1}{ \sqrt{5} } \bigg)^2}{1 -\bigg(\dfrac{1}{ \sqrt{5} } \bigg)^2}\\\\\\:\implies\sf\dfrac{1 - \dfrac{1}{5} }{1 + \dfrac{1}{5}}\\\\\\:\implies\sf\dfrac{\dfrac{(5 - 1)}{\cancel5}}{\dfrac{(5+1)}{\cancel5}}\\\\\\:\implies\sf\cancel\dfrac{4}{6}\\\\\\:\implies\sf\dfrac{2}{3}$

⠀

$\therefore\large\textsf{value of}\:\boxed{\pink{\sf\dfrac{\csc^{2} (\theta) - \sec^{2}( \theta)}{\csc^{2} (\theta) + \sec^{2}( \theta)} =\dfrac{2}{3}}}$

Answer 2

$\bf{\Huge{\boxed{\tt{\green{ANSWER\::}}}}}$

$\bf{\Large{\underline{\bf{[Given\::]}}}}$

$\sf{\large{tan\theta\:=\:\frac{1}{\sqrt{5} } }}$

$\bf{\Large{\underline{\bf{[To\:Find\::]}}}}$

$\sf{\large{\frac{cosec^{2} \theta-sec^{2}\theta }{cosec^{2} \theta+sec^{2}\theta} }}$

$\bf{\Large{\underline{\tt{\red{Explanation\::}}}}}$

We know that tan Ф is;

$\bf{\Large{\boxed{\sf{tan\theta\:=\:\frac{sin\theta}{cos\theta} \:=\:\frac{\frac{1}{cos\theta} }{\frac{1}{sin\theta} } \:=\:\frac{sec\theta}{cosec\theta} }}}}$

$\leadsto\rm{tan^{2}\theta\:=\:\frac{sec^{2} \theta}{cosec^{2}\theta }=\frac{1}{5} }}$

Now,

$\longmapsto\sf{\large{\frac{cosec^{2}\theta-sec^{2} \theta }{cosec^{2}\theta+sec^{2}\theta } =\frac{1-\frac{sec\theta}{cosec\theta} }{1+\frac{sec\theta}{cosec\theta} } }}}$

$\longmapsto\sf{\large{\frac{1-\frac{1}{5} }{1+\frac{1}{5} } }}$

$\longmapsto\sf{\large{\frac{\frac{5-1}{5} }{\frac{5+1}{5} } }}$

$\longmapsto\sf{\large{\frac{\frac{4}{5} }{\frac{6}{5} } }}$

$\longmapsto\sf{\large{\frac{4}{\cancel{5}} *\frac{\cancel{5}}{6} }}$

$\longmapsto\sf{\large{\cancel{\frac{4}{6}} }}$

$\longmapsto\sf{\red{\frac{2}{3} }}$

Thus,

$\bf{\Large{\boxed{\sf{\large{\frac{cosec^{2} \theta-sec^{2}\theta }{cosec^{2} \theta+sec^{2}\theta} \:=\:\frac{2}{3} }}}}}$