Question

If $\text{cosec}A=4p+\frac{1}{16p}$, then the value of $\text{cosec}A+\cot A$ is

• A. $8p$
• B. $\frac{1}{8p}$
• C. $(-8p)$ or $(\frac{1}{8p})$
• D. $\left(8p\right)$ or $(\frac{1}{8p})$

Answer 1

The correct option is D $\left(8p\right)$ or $(\frac{1}{8p})$

$\text{cosec}A=4p+\frac{1}{16p}$

${\text{cosec}}^{2}A=16p^2+\frac{1}{256p^2}+\frac{1}{2}$

Or

$1+{\cot }^{2}A=16p^2+\frac{1}{256p^2}+\frac{1}{2}$

Or

${\cot }^{2}A=16p^2+\frac{1}{256p^2}-\frac{1}{2}$

Or

${\cot }^{2}A=(4p-\frac{1}{16p}{)}^2$

Or

$\cot A=\pm (4p-\frac{1}{16p})$...(i)

Hence

If $\cot A=(4p-\frac{1}{16p})$

Then

$\text{cosec}A+\cot A$

$=4p+\frac{1}{16p}+(4p-\frac{1}{16p})$

$=8p$ ...(ii)

If $\cot A=-(4p-\frac{1}{16p})$

Then

$\text{cosec}A+\cot A$

$=4p+\frac{1}{16p}-(4p-\frac{1}{16p})$

$=\frac{2}{16p}$

$=\frac{1}{8p}$