sin A = 4/5.find 4 tan A - 5 cos A / sec A + 4 cos A.
pls do step by step
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Answers
Answer:
Answer:
\frac{4\tan\theta-5\cos\theta}{\sec\theta+4\cot\theta}=\frac{1}{2}
secθ+4cotθ
4tanθ−5cosθ
=
2
1
Step-by-step explanation:
Given : \sin\theta=\frac{4}{5}sinθ=
5
4
To find : The value of \frac{4\tan\theta-5\cos\theta}{\sec\theta+4\cot\theta}
secθ+4cotθ
4tanθ−5cosθ
Solution :
\sin\theta=\frac{4}{5}sinθ=
5
4
According to trigonometric properties,
\sin\theta=\frac{4}{5}=\frac{P}{H}sinθ=
5
4
=
H
P
i.e Perpendicular P=4, Hypotenuse H=5
Apply Pythagoras theorem,
B=\sqrt{H^2-P^2}B=
H
2
−P
2
B=\sqrt{5^2-4^2}B=
5
2
−4
2
B=\sqrt{25-16}B=
25−16
B=\sqrt{9}B=
9
B=3B=3
We know,
\cos\theta=\frac{B}{H}=\frac{3}{5}cosθ=
H
B
=
5
3
\tan\theta=\frac{P}{B}=\frac{4}{3}tanθ=
B
P
=
3
4
\cot\theta=\frac{B}{P}=\frac{3}{4}cotθ=
P
B
=
4
3
\sec\theta=\frac{H}{B}=\frac{5}{3}secθ=
B
H
=
3
5
Substitute the value in the expression,
=\frac{4(\frac{4}{3})-5(\frac{3}{5})}{(\frac{5}{3})+4(\frac{3}{4})}=
(
3
5
)+4(
4
3
)
4(
3
4
)−5(
5
3
)
=\frac{\frac{16}{3}-3}{\frac{5}{3}+3}=
3
5
+3
3
16
−3
=\frac{\frac{16-9}{3}}{\frac{5+9}{3}}=
3
5+9
3
16−9
=\frac{\frac{7}{3}}{\frac{14}{3}}=
3
14
3
7
=\frac{7}{3}\times \frac{3}{14}=
3
7
×
14
3
=\frac{1}{2}=
2
1
Therefore, \frac{4\tan\theta-5\cos\theta}{\sec\theta+4\cot\theta}=\frac{1}{2}
secθ+4cotθ
4tanθ−5cosθ
=
2
1
Explanation:
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