Question

If cot x =7/8 , what is the value of cot4x?

Answer 1

We're given  $\cot x=\dfrac{7}{8}.$

We know tangent and cotangent ratios are reciprocals to each other. So,

$\cot x=\dfrac{7}{8}\ \ \implies\ \ \tan x=\dfrac{8}{7}$

But we have to find  $\cot(4x).$

So we just apply the identity,

$\large\boxed{\cot(2\theta)=\dfrac{1-\tan^2\theta}{2\tan\theta}}$

This is just the reciprocal of the identity,

$\large\boxed{\tan(2\theta)=\dfrac{2\tan\theta}{1-\tan^2\theta}}$

So,

$\cot(4x)=\cot(2\cdot2x)\\ \\ \\ =\dfrac{1-\tan^2(2x)}{2\tan(2x)}\\ \\ \\ =\dfrac{1-\left(\frac{2\tan x}{1-\tan^2x}\right)^2}{2\cdot\frac{2\tan x}{1-\tan^2x}}\\ \\ \\ =\dfrac{\frac{(1-\tan^2x)^2-(2\tan x)^2}{(1-\tan^2x)^2}}{\frac{4\tan x}{1-\tan^2x}}\\ \\ \\ =\dfrac{(1-\tan^2x)^2-(2\tan x)^2}{4\tan x(1-\tan^2x)}\\ \\ \\ =\dfrac{\tan^4x-6\tan^2x+1}{4\tan x(1-\tan^2x)}$

Now taking  $\tan x=\dfrac{8}{7},$

$\dfrac{\left(\frac{8}{7}\right)^4-6\left(\frac{8}{7}\right)^2+1}{\frac{32}{7}(1-\left(\frac{8}{7}\right)^2)}\ =\ \dfrac{\frac{4096}{2401}-\frac{384}{49}+1}{\frac{32}{7}-\frac{2048}{343}}\\ \\ \\ =\dfrac{\frac{4096-18816+2401}{2401}}{\frac{10976-14336}{2401}}\ =\bold{\dfrac{12319}{3360}}$

Well, I'm not sure whether this fraction is simplified. This is just the final answer!

Hence Found!