If $tan\ Q=\frac{1}{\sqrt{5}}$, what is the value of: $\frac{cosec^{2}\ Q-\sec^{2}\ Q}{cosec^{2}\ Q+\sec^{2}\ Q}$.

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

To do: To find the value of $\frac{cosec^{2}\ Q-\sec^{2}\ Q}{cosec^{2}\ Q+\sec^{2}\ Q}$.

Solution:

As given, $tan\ Q=\frac{1}{\sqrt{5}}$

$\Rightarrow \frac{sin\ Q}{cos\ Q}=\frac{1}{\sqrt{5}}$     [$\because tan\ Q=\frac{sin\ Q}{cos\ Q}$]

$\Rightarrow \frac{\frac{1}{cosec\ Q}}{\frac{1}{sec\ Q}}=\frac{1}{\sqrt{5}}$     [$\because sin\ Q=\frac{1}{cosec\ Q}$ and $cos\ Q=\frac{1}{sec\ Q}$]

$\Rightarrow \frac{sec\ Q}{cosec\ Q}=\frac{1}{\sqrt{5}}$

$\Rightarrow ( \frac{sec\ Q}{cosec\ Q})^2=( \frac{1}{\sqrt{5}})^2$     [On squaring both sides]

$\Rightarrow \frac{sec^2\ Q}{cosec^2\ Q}=\frac{1}{5}\ ........\ ( i)$

Now, $\frac{cosec^{2}\ Q-\sec^{2}\ Q}{cosec^{2}\ Q+\sec^{2}\ Q}$

$\frac{\frac{cosec^{2}\ Q}{cosec^{2}\ Q}-\frac{\sec^{2}\ Q}{cosec^{2}\ Q}}{\frac{cosec^{2}\ Q}{cosec^{2}\ Q}+\frac{\sec^{2}\ Q}{cosec^{2}\ Q}}$  [on dividing both numerators and denominators both by $cosec^{2}\ Q$]

$=\frac{1-\frac{sec^2\ Q}{cosec^{2}\ Q}}{1+\frac{sec^2\ Q}{cosec^{2}\ Q}}$

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

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

$=\frac{\frac{4}{5}}{\frac{6}{5}}$

$=\frac{4}{5}\times\frac{5}{6}$

$=\frac{4}{6}$

$=\frac{2}{3}$

Therefore, $\frac{cosec^{2}\ Q-\sec^{2}\ Q}{cosec^{2}\ Q+\sec^{2}\ Q}=\frac{2}{3}$

Tutorialspoint

Updated on: 10-Oct-2022

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