Question

If ${\tan }^{3}A+{\cot }^{3}A=18$, then the value of ${\tan }^{4}A+{\cot }^{4}A-40$ is equal to

Answer 1

We know that

$a^3+b^3={(a+b)}^3-3(a+b)+a\times b$

Now let $a=\tan A$ and $b=\cot A$

So, we have

${\tan }^{3}A+{\cot }^{3}A={(\tan A+\cot A)}^3-3(\tan A+\cot A).\tan A.\cot A$

${(\tan A+\cot A)}^3-3(\tan A+\cot A).1$

$[\because \tan A.\cot A=1]$

Let$\tan A+\cot A=P$

$\Rightarrow {\tan }^{3}A+{\cot }^{3}A=P^3-3P$

Now, we have $\Rightarrow {\tan }^{3}A+{\cot }^{3}A=18$

$P^3-3P=18$

${\Rightarrow P}^3-3P-18=0$

$\Rightarrow$ we tell that $p=3$ is a solution to above example by hit and trial method.

$\Rightarrow (p-3)(p^2+3p+6)=a\dots ..1$

Now , $p^2+3p+6=0$ has imaginary root for $p$

But $p=\tan A+\cot A$

$\Rightarrow p=$ real for all permissible value of $A$.

Hence, as $p\ne$ imaginary,

$p^2+3p+6\ne 0$

Hence, for equation $1$ to hold $p-3=0\Rightarrow p=3$

Now , ${\tan }^{4}A+{\cot }^{4}A-40={({\tan }^{2}A+{\cot }^{2}A)}^2-2$

$={({\tan }^{2}A+{\cot }^{2}A)}^2-2-40$

$={({({\tan }^{2}A+{\cot }^{2}A)}^2-2\times \tan A\times \cot A)}^2-42$

$={(p^2-2)}^2-42$

$={(3^2-2)}^2-42$.....from equation 2

$={(9-2)}^2-42=7^2-42=7$

So, the answer is $7$