Question

How do you simplify $\dfrac{1+{{\tan }^{2}}x}{1+{{\cot }^{2}}x}$ ?

Answer 1

Hint: Here in this question, we have to use trigonometric fundamental identities. We will use:
$\Rightarrow 1+{{\tan }^{2}}x={{\sec }^{2}}x$
$\Rightarrow 1+{{\cot }^{2}}x=\cos e{{c}^{2}}x$
After applying these identities, we can write ${{\sec }^{2}}x$ as $\dfrac{1}{{{\cos }^{2}}x}$ and ${{\operatorname{cosec}}^{2}}x$as $\dfrac{1}{{{\sin }^{2}}x}$. On dividing these two values, we will obtain the answer.

Complete step by step answer:
Let’s solve the question now.
As we are already aware of the basic functions of trigonometry. Those are:
$\Rightarrow $ Sine (sin)
$\Rightarrow $Cosine (cos)
$\Rightarrow $Tangent (tan)
We are also aware of derived functions as well. They are:
$\Rightarrow $cosec$\theta $ = $\dfrac{1}{\sin \theta }$
$\Rightarrow $sec$\theta $ = $\dfrac{1}{\cos \theta }$
$\Rightarrow $tan$\theta $ = $\dfrac{\sin \theta }{\cos \theta }$ = $\dfrac{1}{\cot \theta }$
$\Rightarrow $cot$\theta $ = $\dfrac{1}{\tan \theta }$ = $\dfrac{\cos \theta }{\sin \theta }$
We can also obtain some values by reciprocating the functions:
$\Rightarrow $sinx = $\dfrac{1}{\text{cosecx}}$ or cosecx = $\dfrac{1}{\sin x}$
$\Rightarrow $cosx = $\dfrac{1}{\sec x}$ or secx = $\dfrac{1}{\cos x}$
$\Rightarrow $tanx = $\dfrac{1}{\cot x}$ or cotx = $\dfrac{1}{\tan x}$
Let’s see some Pythagorean identities as well:
$\Rightarrow {{\cos }^{2}}x+{{\sin }^{2}}x=1$
$\Rightarrow 1+{{\tan }^{2}}x={{\sec }^{2}}x$
$\Rightarrow 1+{{\cot }^{2}}x=\cos e{{c}^{2}}x$
You should know how to derive $1+{{\tan }^{2}}x={{\sec }^{2}}x$.
First, write:
$\Rightarrow 1+{{\tan }^{2}}x$
Now, replace ${{\tan }^{2}}x$ with $\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}$ because tanx can be written in the form of $\dfrac{\sin x}{\cos x}$.
$\Rightarrow 1+\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}$
Now, make the denominator common:
$\Rightarrow \dfrac{{{\cos }^{2}}x+{{\sin }^{2}}x}{{{\cos }^{2}}x}$
We know from formulae that ${{\cos }^{2}}x+{{\sin }^{2}}x=1$. After replacing the value we will get:
$\Rightarrow \dfrac{1}{{{\cos }^{2}}x}$
From reciprocating functions, we know that secx = $\dfrac{1}{\cos x}$. So replace this also:
$\Rightarrow {{\sec }^{2}}x$
In similar fashion, $1+{{\cot }^{2}}x=\cos e{{c}^{2}}x$can also be obtained.
Now:
$\Rightarrow 1+{{\cot }^{2}}x$
Now, replace ${{\cot }^{2}}x$ with $\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x}$ because cotx can be written in the form of $\dfrac{\cos x}{\sin x}$.
$\Rightarrow 1+\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x}$
Now, make the denominator common:
$\Rightarrow \dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{{{\sin }^{2}}x}$
We know from formulae that ${{\cos }^{2}}x+{{\sin }^{2}}x=1$. After replacing the value we will get:
$\Rightarrow \dfrac{1}{{{\sin }^{2}}x}$
From reciprocating functions, we know that cosecx = $\dfrac{1}{\sin x}$. So replace this also:
$\Rightarrow \cos e{{c}^{2}}x$

Now, write the question below.
$\Rightarrow \dfrac{1+{{\tan }^{2}}x}{1+{{\cot }^{2}}x}$
As we know that:
$\Rightarrow 1+{{\tan }^{2}}x={{\sec }^{2}}x$
$\Rightarrow 1+{{\cot }^{2}}x=\cos e{{c}^{2}}x$
So these values will be replaced with new values in the expression, we will get:
$\Rightarrow \dfrac{{{\sec }^{2}}x}{\cos e{{c}^{2}}x}$
We also know that ${{\sec }^{2}}x$ can be written as $\dfrac{1}{{{\cos }^{2}}x}$ and ${{\operatorname{cosec}}^{2}}x$ can be written as $\dfrac{1}{{{\sin }^{2}}x}$. On replacing the values, we will get:
$\Rightarrow \dfrac{\dfrac{1}{{{\cos }^{2}}x}}{\dfrac{1}{{{\sin }^{2}}x}}$
Now, reciprocate the denominator, we will get:
$\Rightarrow \dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}$
As $\dfrac{\sin x}{\cos x}$ forms tanx, so will convert the expression in tangent i.e. tan:

$\therefore {{\tan }^{2}}x$ is the final answer.

Note: If you know all the basic formulae, you can be able to derive any identity very easily. Derived identities should also be learnt which will help you to solve longer questions in less time. In this question, if you don’t know the direct formula for $1+{{\tan }^{2}}x$ and $1+{{\cot }^{2}}x$, you can change them into basic trigonometric functions like sin and cos to obtain the final answer.