Question

If cotx=5/12 then since+1/cotx+secx=?

Answer 1

Answer:

$\frac{sinx+1}{cotx+secx}=\frac{12}{5}$

Step-by-step explanation:

Given:

$cotx=\frac{5}{12}$

$cosec^2x=1+cot^2x$

$cosec^2x=1+(\frac{5}{12})^2$

$cosec^2x=1+\frac{25}{144}$

$cosec^2x=\frac{144+25}{144}$

$cosec^2x=\frac{169}{144}$

$cosecx=\frac{13}{12}$

$sinx=\frac{12}{13}$

$secx=sinx.cotx$

$secx=(\frac{12}{13})(\frac{5}{12})$

$secx=\frac{5}{13}$

Now,

$\frac{sinx+1}{cotx+secx}$

$=\frac{\frac{12}{13}+1}{\frac{5}{12}+\frac{5}{13}}$

$=\frac{\frac{12+13}{13}}{5(\frac{1}{12}+\frac{1}{13})}$

$=\frac{\frac{25}{13}}{5(\frac{13+12}{12*13})}$

$=\frac{\frac{25}{13}}{(\frac{125}{12*13})}$

$=\frac{25}{13}(\frac{12*13}{125})$

$=\frac{12}{5}$