Question

If $m{{\cos }^{2}}A+n{{\sin }^{2}}A=p,$then find the value of ${{\cot }^{2}}A$.

Answer 1

Use the formula ${{\cot }^{2}}A=\dfrac{{{\cos }^{2}}A}{{{\sin }^{2}}A}$ , to obtain ${{\cot }^{2}}A$. To get the values of ${{\cos }^{2}}A$ and ${{\sin }^{2}}A$, first replace ${{\cos }^{2}}A$ with $1-{{\sin }^{2}}A$ in the given equation and get the value of ${{\sin }^{2}}A$ and then replace ${{\sin }^{2}}A$ with $1-{{\cos }^{2}}A$ in the given equation to get the value of ${{\cos }^{2}}A$.

Complete step by step answer:
Given equation is $m{{\cos }^{2}}A+n{{\sin }^{2}}A=p...............\left( 1 \right)$
And we have to find ${{\cot }^{2}}A$.
We know, ${{\cot }^{2}}A=\dfrac{{{\cos }^{2}}A}{{{\sin }^{2}}A}$.
To find the value of ${{\cot }^{2}}A$, we have to find the values of ${{\cos }^{2}}A$ and ${{\sin }^{2}}A$.
We know,
$\begin{aligned} & {{\sin }^{2}}A+{{\cos }^{2}}A=1 \\\ & \Rightarrow {{\sin }^{2}}A=1-{{\cos }^{2}}A \\\ & \Rightarrow {{\cos }^{2}}A=1-{{\sin }^{2}}A \\\ \end{aligned}$
On putting ${{\cos }^{2}}A=1-{{\sin }^{2}}A$ in equation (1), we will get,
$\begin{aligned} & m\left( 1-{{\sin }^{2}}A \right)+n{{\sin }^{2}}A=p \\\ & \Rightarrow m-m{{\sin }^{2}}A+n{{\sin }^{2}}A=p \\\ \end{aligned}$
Taking all constants to RHS and terms containing ${{\sin }^{2}}A$ in the LHS, we will get,
$\Rightarrow -m{{\sin }^{2}}A+n{{\sin }^{2}}A=p-m$
Taking ${{\sin }^{2}}A$ common, we will get,
${{\sin }^{2}}A\left( n-m \right)=p-m$
On dividing both sides by (n – m), we will get,
${{\sin }^{2}}A=\left( \dfrac{p-m}{n-m} \right)............\left( 2 \right)$
Putting ${{\sin }^{2}}A=1-{{\cos }^{2}}A$ in equation (1), we will get,
$\begin{aligned} & m{{\cos }^{2}}A+n\left( 1-{{\cos }^{2}}A \right)=p \\\ & \Rightarrow m{{\cos }^{2}}A+n-n{{\cos }^{2}}A=p \\\ \end{aligned}$
Taking all the constant terms to RHS and all the terms containing ${{\cos }^{2}}A$ to LHS, we will get,
$m{{\cos }^{2}}A-n{{\cos }^{2}}A=p-n$
Taking ${{\cos }^{2}}A$ common, we will get,
$\left( m-n \right){{\cos }^{2}}A=p-n$
Dividing both sides by (m – n), we will get,
${{\cos }^{2}}A=\dfrac{p-n}{m-n}..........\left( 3 \right)$
Now, Putting ${{\cos }^{2}}A$ and ${{\sin }^{2}}A$ in the formula of ${{\cot }^{2}}A$, we will get,
$\begin{aligned} & {{\cot }^{2}}A=\dfrac{{{\cos }^{2}}A}{{{\sin }^{2}}A} \\\ & \Rightarrow {{\cot }^{2}}A=\dfrac{\dfrac{p-n}{m-n}}{\dfrac{p-m}{n-m}} \\\ & \Rightarrow {{\cot }^{2}}A=\dfrac{p-n}{{m-n}}\times \dfrac{{n-m}}{p-m} \\\ & \Rightarrow {{\cot }^{2}}A=\dfrac{n-p}{p-m} \\\ \end{aligned}$

Note:
Another method –
Given equation: $m{{\cos }^{2}}A+n{{\sin }^{2}}A=p$
Divide both sides of equation by ${{\sin }^{2}}A$,
$m\dfrac{{{\cos }^{2}}A}{{{\sin }^{2}}A}+n=\dfrac{P}{{{\sin }^{2}}A}$
We know $\dfrac{{{\cos }^{2}}A}{{{\sin }^{2}}A}={{\cot }^{2}}A$
$\Rightarrow m{{\cot }^{2}}A+n=\dfrac{P}{{{\sin }^{2}}A}........\left( 1 \right)$
Now, in given equation, put ${{\cos }^{2}}A=1-{{\sin }^{2}}A\ \left[ As\ {{\sin }^{2}}A+{{\cos }^{2}}A=1 \right]$
$\begin{aligned} & =m\left( 1-{{\sin }^{2}}A \right)+n{{\sin }^{2}}A=P \\\ & =m-m{{\sin }^{2}}A+n{{\sin }^{2}}A=P \\\ & =\left( n-m \right){{\sin }^{2}}A=P-m \\\ & ={{\sin }^{2}}A=\dfrac{P-m}{n-m} \\\ \end{aligned}$
Om putting ${{\sin }^{2}}A=\dfrac{P-m}{n-m}$ in equation (1), we will get,
$\begin{aligned} & m{{\cot }^{2}}A+n=\dfrac{P\left( n-m \right)}{\left( P-m \right)} \\\ & \Rightarrow m{{\cot }^{2}}A=\dfrac{P\left( n-m \right)}{\left( P-m \right)}-n \\\ \end{aligned}$
Taking LCM & subtracting in RHS,
$\begin{aligned} & \Rightarrow m{{\cot }^{2}}A=\dfrac{P\left( n-m \right)-n\left( P-m \right)}{P-m} \\\ & \Rightarrow m{{\cot }^{2}}A=\dfrac{ {Pn}-Pm- {nP}+nm}{P-m} \\\ & \Rightarrow m{{\cot }^{2}}A=\dfrac{nm-Pm}{P-m} \\\ & \Rightarrow m{{\cot }^{2}}A=\dfrac{m\left( n-P \right)}{P-m} \\\ \end{aligned}$
Dividing both sides by m, we will get,
$\Rightarrow {{\cot }^{2}}A=\dfrac{n-P}{P-m}$
Hence, we got ${{\cot }^{2}}A=\dfrac{n-P}{P-m}$.