# If A is in the fourth quadrant and cos A = 5/13 find the value of

13 sin A+5 sec A divided by 5 tan A + 6cosec A

So cosA = $\frac{5}{13}$

So sinA = (1- cos

^{2}A)

^{1/2}= $\frac{12}{13}$

But in fourth quadrant sin is negative , so we will take -12/13

And secA = 13/5

tanA = sinA/cosA = -12/5

cosecA = 1/sinA = -13/12

Hence $\frac{13\mathrm{sin}A+5secA}{5\mathrm{tan}A+6coescA}=\frac{13\times {\displaystyle \frac{-12}{13}}+5\times {\displaystyle \frac{13}{5}}}{5\times {\displaystyle \frac{-12}{5}}+6\times {\displaystyle \frac{-13}{12}}}=\frac{-2}{37}$

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