Question

Find the value of $M=\cot {{16}^{o}}\cot {{44}^{o}}+\cot {{44}^{o}}\cot {{76}^{o}}-\cot {{76}^{o}}\cot {{16}^{o}}$.

Answer 1

Hint: Observe the relation between the angles of the terms of the given expression. Use trigonometric identities of $\cot \left( A+B \right)=\dfrac{\cot A\cot B-1}{\cot B+\cot A}$ and $\cot \left( A-B \right)=\dfrac{\cot A\cot B+1}{\cot B-\cot A}$ to solve it further.

Complete step-by-step answer:
The given expression is $\cot {{16}^{o}}\cot {{44}^{o}}+\cot {{44}^{o}}\cot {{76}^{o}}-\cot {{76}^{o}}\cot {{16}^{o}}=?$
Let us suppose the value of given expression be M. Hence, we can write
$M=\cot {{16}^{o}}\cot {{44}^{o}}+\cot {{44}^{o}}\cot {{76}^{o}}-\cot {{76}^{o}}\cot {{16}^{o}}$ ………………… (i)
We know the identities of $\cot \left( A+B \right)$ and $\cot \left( A-B \right)$ can be given as
$\cot \left( A+B \right)=\dfrac{\cot A\cot B-1}{\operatorname{cotB}+cotA}$ ………………… (ii)
$\cot \left( A-B \right)=\dfrac{\cot A\cot B+1}{\cot B-\cot A}$…………………(iii)
Now, we can observe the equation (i) and get that the sum of angles of first two expressions are ${{60}^{o}}$ and ${{120}^{o}}$ respectively and difference of angles of last expression $\left( \cot {{76}^{0}}\cot {{16}^{o}} \right)$ is ${{60}^{o}}$.
Hence, we can get value of $cot {{16}^{o}}cot {{44}^{o}}$ from equation (ii) by substituting $A={{16}^{o}}$ and $B={{44}^{o}}$. Hence, we get
$\begin{aligned} & \cot \left( {{44}^{o}}+{{16}^{o}} \right)=\dfrac{\cot {{44}^{o}}\cot {{16}^{o}}-1}{\cot {{44}^{o}}+\cot {{16}^{o}}} \\\ & \cot {{60}^{o}}=\dfrac{\cot {{44}^{o}}\cot {{16}^{o}}-1}{\cot {{44}^{o}}+\cot {{16}^{o}}} \\\ \end{aligned}$
Now, putting value of $\cot {{60}^{o}}$ as $\dfrac{1}{\sqrt{3}}$ and then cross multiplying the above equation, we get
$\begin{aligned} & \dfrac{1}{\sqrt{3}}\left( \cot {{44}^{o}}+\cot {{16}^{o}} \right)=\cot {{44}^{o}}\cot {{16}^{o}}-1 \\\ & \Rightarrow \cot {{44}^{o}}\cot {{16}^{o}}=1+\dfrac{1}{\sqrt{3}}\left( \cot {{44}^{o}}+\cot {{16}^{o}} \right)............(iv) \\\ \end{aligned}$
Similarly put $A={{44}^{o}}$ and $B={{76}^{o}}$ in equation (ii) to get value of $\cot {{44}^{o}}\cot {{76}^{o}}$. So, we get
$\begin{aligned} & \cot \left( {{44}^{o}}+{{76}^{o}} \right)=\dfrac{\cot {{44}^{o}}\cot {{76}^{o}}-1}{\cot {{44}^{o}}+\cot {{76}^{o}}} \\\ & \cot {{120}^{o}}=\dfrac{\cot {{44}^{o}}\cot {{76}^{o}}-1}{\cot {{44}^{o}}+\cot {{76}^{o}}} \\\ \end{aligned}$
Now, we can use the identity $\cot \left( 180-\theta \right)=-\cot \theta$ with $\cot {{120}^{o}}$. Hence, we get
$\begin{aligned} & \cot \left( {{180}^{o}}-{{60}^{o}} \right)=\dfrac{\cot {{44}^{o}}\cot {{76}^{o}}-1}{\cot {{44}^{o}}+\cot {{76}^{o}}} \\\ & -\cot {{60}^{o}}=\dfrac{\cot {{44}^{o}}\cot {{76}^{o}}-1}{\cot {{44}^{o}}+\cot {{76}^{o}}} \\\ \end{aligned}$
Now, put the value of $\cot {{60}^{o}}$as $\dfrac{1}{\sqrt{3}}$and cross-multiply the equation. Hence, we get
$\begin{aligned} & -\dfrac{1}{\sqrt{3}}\left( \cot {{44}^{o}}+\cot {{76}^{o}} \right)=\cot {{44}^{o}}\cot {{76}^{o}}-1 \\\ & \Rightarrow \cot {{44}^{o}}\cot {{76}^{o}}=\dfrac{-1}{\sqrt{3}}\left( \cot {{44}^{o}}+\cot {{76}^{o}} \right)+1...........(v) \\\ \end{aligned}$
And now, put $A={{76}^{o}}$ and $B={{16}^{0}}$in the equation (iii) to get the value of $\cot {{76}^{o}}\cot {{16}^{o}}$. Hence, we get
$\begin{aligned} & \cot \left( {{76}^{o}}-{{16}^{o}} \right)=\dfrac{\cot {{76}^{o}}\cot {{16}^{o}}+1}{\cot {{16}^{o}}-\cot {{76}^{o}}} \\\ & \cot {{60}^{o}}=\dfrac{\cot {{76}^{o}}\cot {{16}^{o}}+1}{\cot {{16}^{o}}-\cot {{76}^{o}}} \\\ \end{aligned}$
Now, put value of $\cot {{60}^{o}}$ and cross-multiply the above equation, hence, we get
$\begin{aligned} & \dfrac{1}{\sqrt{3}}\left( \cot {{16}^{o}}-\cot {{76}^{o}} \right)=\cot 76\cot {{16}^{o}}+1 \\\ & \Rightarrow \cot {{76}^{o}}\cot {{16}^{o}}=\dfrac{1}{\sqrt{3}}\left( \cot {{16}^{o}}-\cot {{76}^{o}} \right)-1............(vi) \\\ \end{aligned}$
Now, put the values of $\cot {{44}^{o}}\cot {{16}^{o}},\cot {{44}^{o}}\cot {{76}^{o}}$ and $\cot {{76}^{o}}\cot {{16}^{o}}$from the equation (iv), (v), (vi) respectively in the equation (i). hence, we get
$\begin{aligned} & M=1+\dfrac{1}{\sqrt{3}}\left( \cot {{44}^{o}}+\cot {{16}^{o}} \right)-\dfrac{1}{\sqrt{3}}\left( \cot {{44}^{o}}+\cot {{76}^{o}} \right)+1-\left( \dfrac{1}{\sqrt{3}}\left( \cot {{16}^{o}}-\cot {{76}^{o}} \right)-1 \right) \\\ & M=1+\dfrac{1}{\sqrt{3}}\cot {{44}^{o}}+\dfrac{1}{\sqrt{3}}\cot {{16}^{o}}-\dfrac{1}{\sqrt{3}}\cot {{44}^{o}}-\dfrac{1}{\sqrt{3}}\cot {{76}^{o}}+1-\dfrac{1}{\sqrt{3}}\cot {{16}^{o}}+\dfrac{1}{\sqrt{3}}\cot {{76}^{o}}+1 \\\ \end{aligned}$
Cancelling the like terms, we get
M = 1+1+1
So, value of $\cot {{16}^{o}}\cot {{44}^{o}}+\cot {{44}^{o}}\cot {{76}^{o}}-\cot {{76}^{o}}\cot {{16}^{o}}$ is 3.

Note: One may think why we did not calculate the value of all the terms in M with a single identity of $\cot \left( A-B \right)$ or $\cot \left( A+B \right)$. We need to observe that, we know the value of $\cot {{60}^{o}}$or $\cot {{120}^{o}}$.
That’s why we split the terms with the required identities.
Another approach for the given question would be that we can put $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ and get expression as
$\begin{aligned} & \dfrac{\cos {{16}^{o}}\cos {{44}^{o}}}{\sin {{16}^{o}}\sin {{44}^{o}}}+\dfrac{\cos {{44}^{o}}\cos {{76}^{o}}}{\sin {{44}^{o}}\sin {{76}^{0}}}-\dfrac{\cos {{76}^{o}}\cos {{16}^{o}}}{\sin {{76}^{o}}\sin {{16}^{o}}} \\\ & =\dfrac{\cos {{16}^{o}}\cos {{44}^{o}}\sin {{76}^{o}}+\cos {{44}^{o}}\cos {{76}^{o}}\sin {{16}^{o}}+\cos {{76}^{o}}\cos {{16}^{o}}\sin {{44}^{o}}}{\sin {{16}^{o}}\sin {{44}^{o}}\sin {{76}^{o}}} \\\ \end{aligned}$
Now multiply by 2 in numerator and denominator and use formula of, $2\cos A\cos B$, in numerator and simplify the denominator by identity
$\sin x\sin \left( {{60}^{o}}-x \right)\sin \left( {{60}^{o}}+x \right)=\dfrac{1}{4}\sin 3x$
Where put $x={{16}^{o}}$, you will get the denominator.
Simplify now to get an answer.