Question

For a given value A, if $\tan A+\cot A=4$, then ${{\tan }^{2}}A+{{\cot }^{2}}A$ is equal to
A. 10
B. 11
C. 12
D. 14

Answer 1

Hint: To solve this question, we have to use the relation of $\cot A=\dfrac{1}{\tan A}$ in the equation $\tan A+\cot A=4$ and we have to solve for $\tan A$ using the quadratic equation concept. After getting the value of $\tan A$, we can get the value of $\cot A$, and then we can substitute the values in ${{\tan }^{2}}A+{{\cot }^{2}}A$ to get the required answer.

Complete step-by-step solution:
In the question, it is given that $\tan A+\cot A=4\to \left( 1 \right)$
We know that $\cot A=\dfrac{1}{\tan A}$
Substituting in the equation-1, we get
$\tan A+\dfrac{1}{\tan A}=4$
Taking L.C.M and multiplying by $\tan A$, we get
$\begin{align}
  & \dfrac{\tan A\times \tan A+1}{\tan A}=4 \\
 & \dfrac{{{\tan }^{2}}A+1}{\tan A}=4 \\
 & {{\tan }^{2}}A+1=4\tan A \\
\end{align}$
Subtracting $4\tan A$ on both sides, we get
${{\tan }^{2}}A-4\tan A+1=0\to \left( 2 \right)$
We can infer that equation-2 is a quadratic equation in $\tan A$.
The roots of a quadratic equation $a{{x}^{2}}+bx+c=0$ are given by
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Comparing this with the equation-2, we get
$\begin{align}
  & x=\tan A \\
 & a=1,b=-4,c=1 \\
\end{align}$
So we can write the values of $\tan A$ as
$\begin{align}
  & \tan A=\dfrac{-\left( -4 \right)\pm \sqrt{{{\left( -4 \right)}^{2}}-4\times 1\times 1}}{2\times 1} \\
 & \tan A=\dfrac{4\pm \sqrt{16-4}}{2}=\dfrac{4\pm \sqrt{12}}{2}=\dfrac{4\pm \sqrt{4\times 3}}{2}=\dfrac{4\pm 2\sqrt{3}}{2} \\
\end{align}$
Cancelling 2 in the numerator and denominator, we get
$\tan A=2\pm \sqrt{3}$
Let us consider the case -1 as $\tan A=2+\sqrt{3}$
As we know that $\cot A=\dfrac{1}{\tan A}$
$\cot A=\dfrac{1}{2+\sqrt{3}}$
Multiplying by $2-\sqrt{3}$ in the numerator and denominator, we get
$\cot A=\dfrac{1}{2+\sqrt{3}}\times \dfrac{2-\sqrt{3}}{2-\sqrt{3}}=\dfrac{2-\sqrt{3}}{\left( 2+\sqrt{3} \right)\times \left( 2-\sqrt{3} \right)}$
The denominator is in the form of $\left( a+b \right)\times \left( a-b \right)={{a}^{2}}-{{b}^{2}}$
Using the relation in the denominator, we get
$\cot A=\dfrac{2-\sqrt{3}}{{{2}^{2}}-{{\left( \sqrt{3} \right)}^{2}}}=\dfrac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$
Using the values of $\tan A\text{ and }\cot A$ in the required expression ${{\tan }^{2}}A+{{\cot }^{2}}A$, we get
${{\tan }^{2}}A+{{\cot }^{2}}A={{\left( 2+\sqrt{3} \right)}^{2}}+{{\left( 2-\sqrt{3} \right)}^{2}}$
The R.H.S is of the form ${{\left( a+b \right)}^{2}}+{{\left( a-b \right)}^{2}}=2\times \left( {{a}^{2}}+{{b}^{2}} \right)$
Applying this in the above R.H.S, we get
${{\left( 2+\sqrt{3} \right)}^{2}}+{{\left( 2-\sqrt{3} \right)}^{2}}=2\times \left( {{2}^{2}}+{{\left( \sqrt{3} \right)}^{2}} \right)=2\times \left( 4+3 \right)=2\times 7=14$
So, for case -1 we get,
${{\tan }^{2}}A+{{\cot }^{2}}A=14$
Let us consider the case -2 as $\tan A=2-\sqrt{3}$
As we know that $\cot A=\dfrac{1}{\tan A}$
$\cot A=\dfrac{1}{2-\sqrt{3}}$
Multiplying by $2+\sqrt{3}$ in the numerator and denominator, we get
$\cot A=\dfrac{1}{2-\sqrt{3}}\times \dfrac{2+\sqrt{3}}{2+\sqrt{3}}=\dfrac{2+\sqrt{3}}{\left( 2-\sqrt{3} \right)\times \left( 2+\sqrt{3} \right)}$
The denominator is in the form of $\left( a+b \right)\times \left( a-b \right)={{a}^{2}}-{{b}^{2}}$
Using the relation in the denominator, we get
$\cot A=\dfrac{2+\sqrt{3}}{{{2}^{2}}-{{\left( \sqrt{3} \right)}^{2}}}=\dfrac{2+\sqrt{3}}{4-3}=2+\sqrt{3}$
Using the values of $\tan A\text{ and }\cot A$ in the required expression ${{\tan }^{2}}A+{{\cot }^{2}}A$, we get
${{\tan }^{2}}A+{{\cot }^{2}}A={{\left( 2-\sqrt{3} \right)}^{2}}+{{\left( 2+\sqrt{3} \right)}^{2}}$
The R.H.S is of the form ${{\left( a-b \right)}^{2}}+{{\left( a+b \right)}^{2}}=2\times \left( {{a}^{2}}+{{b}^{2}} \right)$
Applying this in the above R.H.S, we get
${{\left( 2-\sqrt{3} \right)}^{2}}+{{\left( 2+\sqrt{3} \right)}^{2}}=2\times \left( {{2}^{2}}+{{\left( \sqrt{3} \right)}^{2}} \right)=2\times \left( 4+3 \right)=2\times 7=14$
So, for case -2 we get,
${{\tan }^{2}}A+{{\cot }^{2}}A=14$
From the two cases, we got the same answer, so we can conclude that
${{\tan }^{2}}A+{{\cot }^{2}}A=14$
$\therefore $ ${{\tan }^{2}}A+{{\cot }^{2}}A=14$. The answer is option-D

Note: We can do the question in an alternative way that is to square on both sides of $\tan A+\cot A=4$. Squaring on both sides, we get
$\begin{align}
  & {{\left( \tan A+\cot A \right)}^{2}}={{4}^{2}} \\
 & {{\tan }^{2}}A+{{\cot }^{2}}A+2\times \tan A\times \cot A=16 \\
\end{align}$
As $\cot A=\dfrac{1}{\tan A}$
$\tan A\times \cot A=1$
Using this in the above equation, we get
$\begin{align}
  & {{\tan }^{2}}A+{{\cot }^{2}}A+2\times 1=16 \\
 & {{\tan }^{2}}A+{{\cot }^{2}}A=16-2 \\
 & {{\tan }^{2}}A+{{\cot }^{2}}A=14 \\
\end{align}$
The answer is option-D.