Question

If ${\mathrm{cosec}}^6\left(\theta \right)-{\cot }^6\left(\theta \right)=\mathrm{acot}\left(\mathrm{\theta }\right)+{\mathrm{bcot}}^2\left(\mathrm{\theta }\right)+\mathrm{c},$ then $a+b+c=$

Answer 1

Find the required value

The given expression is

${\mathrm{cosec}}^6\left(\theta \right)-{\cot }^6\left(\theta \right)=\mathrm{acot}\left(\mathrm{\theta }\right)+{\mathrm{bcot}}^2\left(\mathrm{\theta }\right)+\mathrm{c},$

We know that

$a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$

$$\begin{gathered} \therefore L.H.S={\mathrm{cosec}}^6\left(\theta \right)-{\cot }^6\left(\theta \right) \\ =\left({\mathrm{cosec}}^2\left(\theta \right)-{\cot }^2\left(\theta \right)\right)\left({\mathrm{cosec}}^4\left(\theta \right)+{\mathrm{cosec}}^2\left(\theta \right){\cot }^2\left(\theta \right)+{\cot }^4\left(\theta \right)\right) \\ ={\mathrm{cosec}}^4\left(\theta \right)+{\mathrm{cosec}}^2\left(\theta \right){\cot }^2\left(\theta \right)+{\cot }^4\left(\theta \right)\left[\because {\mathrm{cosec}}^2\left(\theta \right)-{\cot }^2\left(\theta \right)=1\right] \\ ={\mathrm{cosec}}^2\left(\theta \right)\left({\mathrm{cosec}}^2\left(\theta \right)+{\cot }^2\left(\theta \right)\right)+{\cot }^4\left(\theta \right) \\ =\left(1+{\cot }^2\left(\theta \right)\right)\left(1+2{\cot }^2\left(\theta \right)\right)+{\cot }^4\left(\theta \right)\left[\because {\mathrm{cosec}}^2\left(\theta \right)-{\cot }^2\left(\theta \right)=1\right] \\ =1+2{\cot }^2\left(\theta \right)+{\cot }^2\left(\theta \right)+2{\cot }^4\left(\theta \right)+{\cot }^4\left(\theta \right) \\ =1+3{\cot }^2\left(\theta \right)+3{\cot }^4\left(\theta \right) \\ R.H.S=a{\cot }^4\left(\theta \right)+b{\cot }^2\left(\theta \right)+c \end{gathered}$$

Comparing $L.H.S$ and $R.H.S$ we get

$$\begin{gathered} a=3,b=3c=1 \\ \therefore a+b+c=7 \end{gathered}$$

Hence, the value of $a+b+c$ is $7$