# Prove the following trigonometric equation:

$\dfrac{{{\tan }^{2}}A}{1+{{\tan }^{2}}A}+\dfrac{{{\cot }^{2}}A}{1+{{\cot }^{2}}A}=1$.

Answer

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Hint: The given question is related to trigonometric identities. Here use the formulae related to the relation between tangent and cotangent of an angle.

Complete step-by-step answer:

We need to prove that the value of $\dfrac{{{\tan }^{2}}A}{1+{{\tan }^{2}}A}+\dfrac{{{\cot }^{2}}A}{1+{{\cot }^{2}}A}$ is equal to $1$ .

First, we will consider the left-hand side of the equation which is given as $\dfrac{{{\tan }^{2}}A}{1+{{\tan }^{2}}A}+\dfrac{{{\cot }^{2}}A}{1+{{\cot }^{2}}A}$. We will take the LCM of the denominators and solve the fraction to determine the value of the left-hand side of the equation. The LCM of the denominators in the left-hand side of the equation is given as $\left( 1+{{\tan }^{2}}A \right)\left( 1+{{\cot }^{2}}A \right)$ .

So, the left-hand side of the equation becomes \[\dfrac{{{\tan }^{2}}A\left( 1+{{\cot }^{2}}A \right)}{\left( 1+{{\tan }^{2}}A \right)\left( 1+{{\cot }^{2}}A \right)}+\dfrac{{{\cot }^{2}}A\left( 1+{{\tan }^{2}}A \right)}{\left( 1+{{\tan }^{2}}A \right)\left( 1+{{\cot }^{2}}A \right)}\].

$=\dfrac{{{\tan }^{2}}A\left( 1+{{\cot }^{2}}A \right)+{{\cot }^{2}}A\left( 1+{{\tan }^{2}}A \right)}{\left( 1+{{\tan }^{2}}A \right)\left( 1+{{\cot }^{2}}A \right)}$

Now, we will open the brackets and multiply the terms in the denominator. So, we get:

$\dfrac{{{\tan }^{2}}A}{1+{{\tan }^{2}}A}+\dfrac{{{\cot }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{{{\tan }^{2}}A+{{\tan }^{2}}A{{\cot }^{2}}A+{{\cot }^{2}}A+{{\tan }^{2}}A{{\cot }^{2}}A}{1+{{\tan }^{2}}A+{{\tan }^{2}}A{{\cot }^{2}}A+{{\cot }^{2}}A}....(i)$

Now, we know that $\cot A=\dfrac{1}{\tan A}$ . So, $\cot A\tan A=1$ . Hence, ${{\cot }^{2}}A{{\tan }^{2}}A=1$.

Now, we will substitute ${{\cot }^{2}}A{{\tan }^{2}}A=1$ in equation $(i)$. So, we get: $\dfrac{{{\tan }^{2}}A}{1+{{\tan }^{2}}A}+\dfrac{{{\cot }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{{{\tan }^{2}}A+1+{{\cot }^{2}}A+1}{1+{{\tan }^{2}}A+{{\cot }^{2}}A+1}$. Clearly, the values of the numerator and the denominator are the same. So, the value of the fraction will be equal to $1$ . So, the value of $\dfrac{{{\tan }^{2}}A}{1+{{\tan }^{2}}A}+\dfrac{{{\cot }^{2}}A}{1+{{\cot }^{2}}A}$ will be equal to $1$, i.e.

$\dfrac{{{\tan }^{2}}A}{1+{{\tan }^{2}}A}+\dfrac{{{\cot }^{2}}A}{1+{{\cot }^{2}}A}=1$.

Hence, the value of the left-hand side of the equation is equal to $1$.

Now, we will consider the right-hand side of the equation. The value of the right-hand side of the equation is also equal to $1$.

LHS=RHS. Hence, proved.

Note: While taking the LCM and multiplying the terms in the denominator, make sure that no sign mistakes are made. They can result in wrong answers. So, such mistakes should be avoided.

Complete step-by-step answer:

We need to prove that the value of $\dfrac{{{\tan }^{2}}A}{1+{{\tan }^{2}}A}+\dfrac{{{\cot }^{2}}A}{1+{{\cot }^{2}}A}$ is equal to $1$ .

First, we will consider the left-hand side of the equation which is given as $\dfrac{{{\tan }^{2}}A}{1+{{\tan }^{2}}A}+\dfrac{{{\cot }^{2}}A}{1+{{\cot }^{2}}A}$. We will take the LCM of the denominators and solve the fraction to determine the value of the left-hand side of the equation. The LCM of the denominators in the left-hand side of the equation is given as $\left( 1+{{\tan }^{2}}A \right)\left( 1+{{\cot }^{2}}A \right)$ .

So, the left-hand side of the equation becomes \[\dfrac{{{\tan }^{2}}A\left( 1+{{\cot }^{2}}A \right)}{\left( 1+{{\tan }^{2}}A \right)\left( 1+{{\cot }^{2}}A \right)}+\dfrac{{{\cot }^{2}}A\left( 1+{{\tan }^{2}}A \right)}{\left( 1+{{\tan }^{2}}A \right)\left( 1+{{\cot }^{2}}A \right)}\].

$=\dfrac{{{\tan }^{2}}A\left( 1+{{\cot }^{2}}A \right)+{{\cot }^{2}}A\left( 1+{{\tan }^{2}}A \right)}{\left( 1+{{\tan }^{2}}A \right)\left( 1+{{\cot }^{2}}A \right)}$

Now, we will open the brackets and multiply the terms in the denominator. So, we get:

$\dfrac{{{\tan }^{2}}A}{1+{{\tan }^{2}}A}+\dfrac{{{\cot }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{{{\tan }^{2}}A+{{\tan }^{2}}A{{\cot }^{2}}A+{{\cot }^{2}}A+{{\tan }^{2}}A{{\cot }^{2}}A}{1+{{\tan }^{2}}A+{{\tan }^{2}}A{{\cot }^{2}}A+{{\cot }^{2}}A}....(i)$

Now, we know that $\cot A=\dfrac{1}{\tan A}$ . So, $\cot A\tan A=1$ . Hence, ${{\cot }^{2}}A{{\tan }^{2}}A=1$.

Now, we will substitute ${{\cot }^{2}}A{{\tan }^{2}}A=1$ in equation $(i)$. So, we get: $\dfrac{{{\tan }^{2}}A}{1+{{\tan }^{2}}A}+\dfrac{{{\cot }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{{{\tan }^{2}}A+1+{{\cot }^{2}}A+1}{1+{{\tan }^{2}}A+{{\cot }^{2}}A+1}$. Clearly, the values of the numerator and the denominator are the same. So, the value of the fraction will be equal to $1$ . So, the value of $\dfrac{{{\tan }^{2}}A}{1+{{\tan }^{2}}A}+\dfrac{{{\cot }^{2}}A}{1+{{\cot }^{2}}A}$ will be equal to $1$, i.e.

$\dfrac{{{\tan }^{2}}A}{1+{{\tan }^{2}}A}+\dfrac{{{\cot }^{2}}A}{1+{{\cot }^{2}}A}=1$.

Hence, the value of the left-hand side of the equation is equal to $1$.

Now, we will consider the right-hand side of the equation. The value of the right-hand side of the equation is also equal to $1$.

LHS=RHS. Hence, proved.

Note: While taking the LCM and multiplying the terms in the denominator, make sure that no sign mistakes are made. They can result in wrong answers. So, such mistakes should be avoided.

Last updated date: 27th May 2023

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