Question

If, $\tan 9{}^\circ =\dfrac{x}{y}$ then, value of $\dfrac{{{\sec }^{2}}81{}^\circ }{1+{{\cot }^{2}}81{}^\circ }$ is
(A). $\dfrac{{{x}^{3}}}{{{y}^{3}}}$
(B). $\dfrac{{{x}^{4}}}{{{y}^{4}}}$
(C). $\dfrac{{{x}^{5}}}{{{y}^{5}}}$
(D). $\dfrac{{{y}^{2}}}{{{x}^{2}}}$

Answer 1

Hint: Take $\dfrac{{{\sec }^{2}}81{}^\circ }{1+{{\cot }^{2}}81{}^\circ }$, use $1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta $. After that, use $\dfrac{1}{\sin \theta }=\csc \theta $ and $\dfrac{1}{\cos \theta }=\sec \theta $.
Simplify it and use $\cot (90{}^\circ -\theta )=\tan \theta $. Try it, you will get the answer.

Complete step-by-step answer:

In question it is given that $\tan 9{}^\circ =\dfrac{x}{y}$ and we have to find the value of $\dfrac{{{\sec }^{2}}81{}^\circ }{1+{{\cot }^{2}}81{}^\circ }$.
We know that, $1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta $,
Now taking $\dfrac{{{\sec }^{2}}81{}^\circ }{1+{{\cot }^{2}}81{}^\circ }=\dfrac{{{\sec }^{2}}81{}^\circ }{{{\csc }^{2}}81{}^\circ }$
We know that, $\dfrac{1}{\sin \theta }=\csc \theta $ and $\dfrac{1}{\cos \theta }=\sec \theta $.
So we get,
$\begin{align}
  & \dfrac{{{\sec }^{2}}81{}^\circ }{1+{{\cot }^{2}}81{}^\circ }=\dfrac{{{\sec }^{2}}81{}^\circ }{{{\csc }^{2}}81{}^\circ } \\
 & \Rightarrow \dfrac{{{\sec }^{2}}81{}^\circ }{1+{{\cot }^{2}}81{}^\circ }=\dfrac{\dfrac{1}{{{\cos }^{2}}81{}^\circ }}{\dfrac{1}{{{\sin }^{2}}81{}^\circ }} \\
\end{align}$
Simplifying we get,
$\dfrac{{{\sec }^{2}}81{}^\circ }{1+{{\cot }^{2}}81{}^\circ }=\dfrac{\dfrac{1}{{{\cos }^{2}}81{}^\circ }}{\dfrac{1}{{{\sin }^{2}}81{}^\circ }}=\dfrac{{{\sin }^{2}}81{}^\circ }{{{\cos }^{2}}81{}^\circ }$
We know $\sin (90{}^\circ -\theta )=cos\theta ,cos(90{}^\circ -\theta )=sin\theta $, so above equation can be written as,
$\dfrac{{{\sec }^{2}}81{}^\circ }{1+{{\cot }^{2}}81{}^\circ }=\dfrac{{{\left[ \sin (90{}^\circ -9{}^\circ )\right]}^{2}}}{{{\left[ \cos (90{}^\circ -9{}^\circ ) \right]}^{2}}}=\dfrac{{{\cos }^{2}}9{}^\circ }{{{\sin}^{2}}9{}^\circ }={\cot}^2 9{}^\circ $

Here, $\tan 9{}^\circ =\dfrac{x}{y}$,
so substituting above we get,
$\dfrac{{{\sec }^{2}}81{}^\circ }{1+{{\cot }^{2}}81{}^\circ }={{\left( \dfrac{1}{\tan 9{}^\circ } \right)}^{2}}={{\left( \dfrac{y}{x} \right)}^{2}}=\dfrac{{{y}^{2}}}{{{x}^{2}}}$
Hence, the value of $\dfrac{{{\sec }^{2}}81{}^\circ }{1+{{\cot }^{2}}81{}^\circ }$ is $\dfrac{{{y}^{2}}}{{{x}^{2}}}$.
The correct answer is option(D).

Note: Read the question carefully. You should be thorough with the concept of trigonometry. While simplifying, do not miss any term. Don’t make silly mistakes. You should know the identities such as $1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta $, $\cot (90{}^\circ -\theta )=\tan \theta $, $\dfrac{1}{\sin \theta }=\csc \theta $etc. These all properties are used in the above problem.
Another approach is directly converting the given expression in terms of $9{}^\circ $ then simplify.
$\dfrac{{{\sec }^{2}}81{}^\circ }{1+{{\cot }^{2}}81{}^\circ }=\dfrac{\dfrac{1}{{{\cos }^{2}}(90{}^\circ -9{}^\circ )}}{1+{{\cot }^{2}}(90{}^\circ -9{}^\circ )}=\dfrac{\dfrac{1}{si{{n}^{2}}(9{}^\circ )}}{1+{{\tan }^{2}}(9{}^\circ )}=\dfrac{cs{{c}^{2}}(9{}^\circ )}{1+{{\tan }^{2}}(9{}^\circ )}$
In this method, you will get the same answer, but it will be tedious.