# If cot x=(2n^2+2mn)/(m^2+2mn) then find cscx?

Mar 9, 2016

$\csc x = \frac{{m}^{2} + 2 m n + 2 {n}^{2}}{{m}^{2} + 2 m n}$

#### Explanation:

If $\cot x = \frac{2 {n}^{2} + 2 m n}{{m}^{2} + 2 m n}$ then prove that

${\csc}^{2} x = 1 + {\left(\frac{2 {n}^{2} + 2 m n}{{m}^{2} + 2 m n}\right)}^{2}$ or

${\csc}^{2} x = \frac{{\left({m}^{2} + 2 m n\right)}^{2} + {\left(2 {n}^{2} + 2 m n\right)}^{2}}{{m}^{2} + 2 m n} ^ 2$

= $\frac{\left({m}^{4} + 4 {m}^{2} {n}^{2} + 4 {m}^{3} n\right) + \left(4 {n}^{4} + 4 {m}^{2} {n}^{2} + 8 m {n}^{3}\right)}{{m}^{2} + 2 m n} ^ 2$

(using identity (a+b)^2=a^2+b^2+2ab)) and now adding and rearranging terms

${\csc}^{2} x = \frac{\left({m}^{4} + 4 {m}^{3} n + 8 {m}^{2} {n}^{2} + 8 m {n}^{3} + 4 {n}^{4}\right)}{{m}^{2} + 2 m n} ^ 2$ ...(A)

Now ${\left({m}^{2} + 2 m n + 2 {n}^{2}\right)}^{2}$ becomes

$\left({m}^{4} + 4 {m}^{2} {n}^{2} + 4 {n}^{4} + 2 \cdot {m}^{2} \cdot 2 m n + 2 \cdot {m}^{2} \cdot 2 {n}^{2} + 2 \cdot 2 m n \cdot 2 {n}^{2}\right)$

(using identity ${\left(a + b + c\right)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2 a b + 2 a c + 2 b c$)

or $\left({m}^{4} + 4 {m}^{2} {n}^{2} + 4 {n}^{4} + 4 {m}^{3} n + 4 {m}^{2} {n}^{2} + 8 m {n}^{3}\right)$

or $\left({m}^{4} + 4 {m}^{3} n + 8 {m}^{2} {n}^{2} + 8 m {n}^{3} + 4 {n}^{4}\right)$ ....(B)

Note that in numerator of (A) and in (B), we have arranged monomials in decreasing degrees of $m$ and increasing degree of $n$. Also note that both are equal.

Hence, ${\csc}^{2} x = \frac{{\left({m}^{2} + 2 m n + 2 {n}^{2}\right)}^{2}}{{\left({m}^{2} + 2 m n\right)}^{2}}$

or $\csc x = \frac{{m}^{2} + 2 m n + 2 {n}^{2}}{{m}^{2} + 2 m n}$