Question

If a is the fourth quadrant and cos a = 5/13 then find the value of 13sin a + 5 sec a / 5 tan a + 6 cosec a​

Answer 1

$\textbf{Given:}$

$\text{ cosA=\dfrac{5}{13} and A is in 4 th quadrant}$

$\textbf{To find:}$

$\text{The value of \dfrac{13\,sinA+5\,secA}{5\,tanA+6\,cosecA} }$

$\textbf{Solution:}$

$\text{Consider,}$

$sin^2A=1-cos^2A$

$sin^2A=1-(\dfrac{5}{13})^2$

$sin^2A=1-\dfrac{25}{169}$

$sin^2A=\dfrac{169-25}{169}$

$sin^2A=\dfrac{144}{169}$

$sinA=\pm\,\dfrac{12}{13}$

$\text{Since A lies in 4 th quadrant, sinA is negative}$

$\implies\,sinA=\dfrac{-12}{13}$

$tanA=\dfrac{sinA}{cosA}$

$=\dfrac{\dfrac{-12}{13}}{\dfrac{5}{13}}$

$=\dfrac{-12}{5}$

$\text{Now}$

$\dfrac{13\,sinA+5\,secA}{5\,tanA+6\,cosecA}$

$=\dfrac{13(\frac{-12}{13})+5(\frac{13}{5})}{5(\frac{-12}{5})+6(\frac{-13}{12})}$

$=\dfrac{-12+13}{-12-\frac{13}{2}}$

$=\dfrac{1}{\frac{-24-13}{2}}$

$=\dfrac{1}{\frac{-37}{2}}$

$=\dfrac{-2}{37}$

$\textbf{Answer:}$

$\textbf{The value of \bf\dfrac{13\,sinA+5\,secA}{5\,tanA+6\,cosecA} is \bf\dfrac{-2}{37} }$

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Answer 2

here's ur answer..

hope it helps u