Given tan θ=15√, what is the value of cosec2 θ−sec2 θcosec2 θ+sec2 θ?
Question:

Given $\tan \theta=\frac{1}{\sqrt{5}}$, what is the value of $\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}$ ?

Solution:

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

We know that: $\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$

$\frac{\text { Perpendicular }}{\text { Base }}=\frac{1}{\sqrt{5}}$

Hypotenuse $=\sqrt{(\text { Perpendicular })^{2}+(\text { Base })^{2}}$

Hypotenuse $=\sqrt{1+5}$

Hypotenuse $=\sqrt{6}$

Now we find, $\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}$

$=\frac{\frac{(\text { hypotenuse })^{2}}{(\text { Perpendicular })^{2}}-\frac{(\text { hypotenuse })^{2}}{(\text { Base })^{2}}}{\frac{(\text { hypotenuse })^{2}}{(\text { Perpendicular })^{2}}+\frac{(\text { hypotenuse })^{2}}{(\text { Base })^{2}}}$

$=\frac{\frac{(\sqrt{6})^{2}}{(1)^{2}}-\frac{(\sqrt{6})^{2}}{(\sqrt{5})^{2}}}{\frac{(\sqrt{6})^{2}}{(1)^{2}}+\frac{(\sqrt{6})^{2}}{(\sqrt{5})^{2}}}$

$=\frac{\frac{6}{1}-\frac{6}{5}}{\frac{6}{1}+\frac{6}{5}}$

$=\frac{\frac{24}{\frac{5}{36}}}{\frac{5}{5}}$

$=\frac{2}{3}$

Hence the value of $\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}$ is $\frac{2}{3}$

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