Question

Is LHS=RHS?
$\frac{1}{cosec\theta +\cot \theta }-\frac{1}{\sin \theta }=\frac{1}{\sin \theta }-\frac{1}{cosec\theta -\cot \theta }$

• A. Yes
• B. No
• C. Can't say
• D. Data insufficient

Answer 1

Answer: (A)

Yes

$LHS$

$=\frac{sinx}{1+cosx}-\frac{1}{sinx}$
$=\frac{si{n}^{2}x-1-cosx}{sinx(1+cosx)}$
$=\frac{-(cosx+co{s}^{2}x)}{sinx(1+cosx)}$
$=\frac{-cos\left(x\right)(1+cosx)}{sinx(1+cos(x\left)\right)}$
$=\frac{-cosx}{sinx}$

$RHS$
$=\frac{1-cosx-si{n}^{2}x}{(1-cosx)\left(sinx\right)}$
$=\frac{co{s}^{2}x-cosx}{sinx(1-cosx)}$
$=\frac{-cosx(1-cosx)}{sinx(1-cosx)}$
$=\frac{-cosx}{sinx}$

$HenceLHS=RHS.$