Question

If $\tan x=\dfrac{1}{\sqrt{7}}$ then $\dfrac{{{\csc }^{2}}\theta -{{\sec }^{2}}\theta }{{{\csc }^{2}}\theta +{{\sec }^{2}}\theta }$ is equal to

a)$\dfrac{5}{7}$
b)$\dfrac{3}{7}$
c)$\dfrac{1}{12}$
d)$\dfrac{3}{4}$

Answer 1

Hint: Here, first we have to draw the figure and with the help of definition$\tan x=\dfrac{opposite\text{ }side}{adjacent\text{ }side}$mark the opposite side and adjacent side. From the figure find the value of hypotenuse using the Pythagoras theorem. Next, find $\sin x$ and $\cos x$. Now, to find $\csc x$ and $\sec x$ we have to apply the formulas:

$\begin{align}

  & \csc x=\dfrac{1}{\sin x} \\

 & \sec x=\dfrac{1}{\cos x} \\

\end{align}$

Complete step-by-step answer:
Here, we are given that $\tan x=\dfrac{1}{\sqrt{7}}$.

Now, we have to find the value of $\dfrac{{{\csc }^{2}}\theta -{{\sec }^{2}}\theta }{{{\csc

}^{2}}\theta +{{\sec }^{2}}\theta }$.

We know that,

$\tan x=\dfrac{opposite\text{ }side}{adjacent\text{ }side}$

We have the figure,

Here, from the figure we can say that,

Opposite side = AC

Adjacent side = AB

Hypotenuse = BC

$\Delta ABC$ is a right angled triangle. Hence, we can apply the Pythagoras theorem.

Now, by Pythagoras theorem we have,

$\begin{align}

  & {{(Hypotenuse)}^{2}}={{(Opposite\text{ }side)}^{2}}+{{(Adjacent\text{ }side)}^{2}} \\

 & \Rightarrow {{(BC)}^{2}}={{(AC)}^{2}}+{{(AB)}^{2}} \\

\end{align}$

Here, we have,

$\tan x=\dfrac{AC}{AB}$

AC = 1

AB = $\sqrt{7}$

Now, we can write:

$\begin{align}

  & {{(BC)}^{2}}={{1}^{2}}+{{\left( \sqrt{7} \right)}^{2}} \\

 & \Rightarrow {{(BC)}^{2}}=1+7 \\

 & \Rightarrow {{(BC)}^{2}}=8 \\

\end{align}$

Next, by taking square root on both the sides we get,

$BC=\sqrt{8}$

We know that,

$\begin{align}

  & \sin x=\dfrac{Opposite\text{ }side}{Hypotenuse} \\

 & \Rightarrow \sin x=\dfrac{AC}{BC} \\

 & \Rightarrow \sin x=\dfrac{1}{\sqrt{8}} \\

\end{align}$

We also have that,

$\begin{align}

  & \csc x=\dfrac{1}{\sin x} \\

 & \Rightarrow \csc x=\dfrac{1}{\dfrac{1}{\sqrt{8}}} \\

 & \Rightarrow \csc x=1\times \dfrac{\sqrt{8}}{1} \\

 & \Rightarrow \csc x=\sqrt{8} \\

\end{align}$

Similarly, we have,

$\begin{align}

  & \cos x=\dfrac{\text{Adjacent }side}{Hypotenuse} \\

 & \Rightarrow \cos x=\dfrac{AB}{BC} \\

 & \Rightarrow \cos x=\dfrac{\sqrt{7}}{\sqrt{8}} \\

\end{align}$

We also know that,

$\begin{align}

  & \sec x=\dfrac{1}{\cos x} \\

 & \Rightarrow \sec x=\dfrac{1}{\dfrac{\sqrt{7}}{\sqrt{8}}} \\

 & \Rightarrow \sec x=1\times \dfrac{\sqrt{8}}{\sqrt{7}} \\

 & \Rightarrow \sec x=\dfrac{\sqrt{8}}{\sqrt{7}} \\

\end{align}$

Next, we have to find $\dfrac{{{\csc }^{2}}\theta -{{\sec }^{2}}\theta }{{{\csc }^{2}}\theta +{{\sec }^{2}}\theta }$.

$\begin{align}

  & \dfrac{{{\csc }^{2}}\theta -{{\sec }^{2}}\theta }{{{\csc }^{2}}\theta +{{\sec }^{2}}\theta }=\dfrac{{{\left( \sqrt{8} \right)}^{2}}-{{\left( \dfrac{\sqrt{8}}{\sqrt{7}} \right)}^{2}}}{{{\left( \sqrt{8} \right)}^{2}}+{{\left( \dfrac{\sqrt{8}}{\sqrt{7}} \right)}^{2}}} \\

 & \Rightarrow \dfrac{{{\csc }^{2}}\theta -{{\sec }^{2}}\theta }{{{\csc }^{2}}\theta +{{\sec }^{2}}\theta }=\dfrac{8-\dfrac{8}{7}}{8+\dfrac{8}{7}} \\

\end{align}$

Now, by taking LCM we get,

$\Rightarrow \dfrac{{{\csc }^{2}}\theta -{{\sec }^{2}}\theta }{{{\csc }^{2}}\theta +{{\sec }^{2}}\theta }=\dfrac{\dfrac{8\times 7-8}{7}}{\dfrac{8\times 7+8}{7}}$

$\Rightarrow \dfrac{{{\csc }^{2}}\theta -{{\sec }^{2}}\theta }{{{\csc }^{2}}\theta +{{\sec }^{2}}\theta }=\dfrac{\dfrac{56-7}{7}}{\dfrac{56+7}{7}}$

$\begin{align}

  & \Rightarrow \dfrac{{{\csc }^{2}}\theta -{{\sec }^{2}}\theta }{{{\csc }^{2}}\theta +{{\sec }^{2}}\theta }=\dfrac{\dfrac{48}{7}}{\dfrac{64}{7}} \\

 & \Rightarrow \dfrac{{{\csc }^{2}}\theta -{{\sec }^{2}}\theta }{{{\csc }^{2}}\theta +{{\sec }^{2}}\theta }=\dfrac{48}{7}\times \dfrac{7}{64} \\

\end{align}$

Next, by cancellation we get:

$\Rightarrow \dfrac{{{\csc }^{2}}\theta -{{\sec }^{2}}\theta }{{{\csc }^{2}}\theta +{{\sec }^{2}}\theta }=\dfrac{48}{64}$

Again by cancellation we obtain,

$\Rightarrow \dfrac{{{\csc }^{2}}\theta -{{\sec }^{2}}\theta }{{{\csc }^{2}}\theta +{{\sec }^{2}}\theta }=\dfrac{3}{4}$

Therefore, we can say that when $\tan x=\dfrac{1}{\sqrt{7}}$ , then $\dfrac{{{\csc }^{2}}\theta -{{\sec }^{2}}\theta }{{{\csc }^{2}}\theta +{{\sec }^{2}}\theta }=\dfrac{3}{4}$.
Hence, the correct answer for this question is option (d).

Note: Here, you should have an idea about the trigonometric ratios. To have a better understanding you have to construct a right triangle and mark the angle as x , it’s opposite side as 1 and it’s adjacent side as $\sqrt{7}$.