If cos A =5/13 and A lies A in 4th quadrant. Find the value of 13sin A+5 sec A÷5tan A+6 cosec A
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Step-by-step explanation:
\textbf{Given:}Given:
\text{$cosA=\dfrac{5}{13}$ and A is in 4 th quadrant}cosA=
13
5
and A is in 4 th quadrant
\textbf{To find:}To find:
\text{The value of $\dfrac{13\,sinA+5\,secA}{5\,tanA+6\,cosecA}$}The value of
5tanA+6cosecA
13sinA+5secA
\textbf{Solution:}Solution:
\text{Consider,}Consider,
sin^2A=1-cos^2Asin
2
A=1−cos
2
A
sin^2A=1-(\dfrac{5}{13})^2sin
2
A=1−(
13
5
)
2
sin^2A=1-\dfrac{25}{169}sin
2
A=1−
169
25
sin^2A=\dfrac{169-25}{169}sin
2
A=
169
169−25
sin^2A=\dfrac{144}{169}sin
2
A=
169
144
sinA=\pm\,\dfrac{12}{13}sinA=±
13
12
\text{Since A lies in 4 th quadrant, $sinA$ is negative}Since A lies in 4 th quadrant, sinA is negative
\implies\,sinA=\dfrac{-12}{13}⟹sinA=
13
−12
tanA=\dfrac{sinA}{cosA}tanA=
cosA
sinA
=\dfrac{\dfrac{-12}{13}}{\dfrac{5}{13}}=
13
5
13
−12
=\dfrac{-12}{5}=
5
−12
\text{Now}Now
\dfrac{13\,sinA+5\,secA}{5\,tanA+6\,cosecA}
5tanA+6cosecA
13sinA+5secA
=\dfrac{13(\frac{-12}{13})+5(\frac{13}{5})}{5(\frac{-12}{5})+6(\frac{-13}{12})}=
5(
5
−12
)+6(
12
−13
)
13(
13
−12
)+5(
5
13
)
=\dfrac{-12+13}{-12-\frac{13}{2}}=
−12−
2
13
−12+13
=\dfrac{1}{\frac{-24-13}{2}}=
2
−24−13
1
=\dfrac{1}{\frac{-37}{2}}=
2
−37
1
=\dfrac{-2}{37}=
37
−2
\textbf{Answer:}Answer:
\textbf{The value of $\bf\dfrac{13\,sinA+5\,secA}{5\,tanA+6\,cosecA}$ is $\bf\dfrac{-2}{37}$}The value of
5tanA+6cosecA
13sinA+5secA
is
37
−2
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