Question

Resolve $\frac{(2+cosx)}{(1-cosx)(3+cosx)(1+cosx{)}^2}$ into partial fractions.

• A. $\frac{3}{8(1-cosx)}-\frac{1}{8(3+cosx)}+\frac{1}{4(1+cosx)}+\frac{4}{4(1+cosx{)}^2}$
• B. $\frac{3}{16(1-cosx)}-\frac{1}{16(3+cosx)}+\frac{1}{8(1+cosx)}+\frac{4}{4(1+cosx{)}^2}$
• C. $\frac{3}{16(1-cosx)}-\frac{1}{16(3+cosx)}+\frac{1}{4(1+cosx)}+\frac{4}{4(1+cosx{)}^2}$
• D. $\frac{3}{16(1-cosx)}-\frac{1}{16(3+cosx)}+\frac{1}{4(1+cosx)}+\frac{4}{8(1+cosx{)}^2}$

Answer 1

Answer: (C)

$\frac{3}{16(1-cosx)}-\frac{1}{16(3+cosx)}+\frac{1}{4(1+cosx)}+\frac{4}{4(1+cosx{)}^2}$

In the given fraction everwhere put $cosx=t$ for the sake of partial fractions.
$\therefore \frac{(2+cosx)}{(1-cosx)(3+cosx)(1+cosx{)}^2}=\frac{(2+t)}{(1-t)(3+t)(1+t{)}^2}$
Let$\frac{(2+t)}{(1-t)(3+t)(1+t{)}^2}=\frac{A}{1-t}+\frac{B}{3+t}+\frac{C}{1+t}=\frac{D}{(1+t{)}^2}$ ...(i)
$\Rightarrow 2+t=A(3+t)(1+t{)}^2+B(1-t\left)\right(1+t{)}^2$

$+C(1-t)(1+t)(3+t)+D(1-t)(3+t)$ ...(ii)
Puttitng $1-t=0$ or $t=1$
In (ii), we obtan $2+1=A(3+1)(1+1{)}^2+0+0+0$
$\therefore A=\frac{3}{16}$
Putting $1+t=0$ or $t=-1$ in (ii), we obtain
$2-1=0+0+0+0+D(1+1)(3-1)$
$\therefore D=\frac{1}{4}$
Putting $3+t=0$ or $t=-3$ in (ii), we obtain
$2-3=0+B(1+3)(1-3{)}^2+0+0$
$\therefore B=\frac{-1}{16}$
Comparing the coefficient of $t^3$ in (ii), we obtain
$0=A-B-C$
$\Rightarrow 0=\frac{3}{16}+\frac{1}{16}-C\Rightarrow C=\frac{1}{4}$
Substituting the values of $A,B,C,D$ in (i), then
$\frac{(2+t)}{(1-t)(3+t)(1+t{)}^2}=\frac{3}{16(1-t)}-\frac{1}{16(3+t)}+\frac{1}{4(1+t)}+\frac{4}{4(1+t{)}^2}$
Putting $t=cosx$ in last then,

$\frac{(2+cosx)}{(1-cosx)(3+cosx)(1+cosx{)}^2}=\frac{3}{16(1-cosx)}-\frac{1}{16(3+cosx)}$

$+\frac{1}{4(1+cosx)}+\frac{4}{4(1+cosx{)}^2}$
which are the required partial fractions.
Hence, option 'C' is correct.