Question

If sin theta = 4/5 , find the value of 4tan theta -5 cos theta/ sec theta plus 4 cot theta

Answer 1

1/2 will be the answer

Answer 2

Answer:

$\frac{4\tan\theta-5\cos\theta}{\sec\theta+4\cot\theta}=\frac{1}{2}$

Step-by-step explanation:

Given : $\sin\theta=\frac{4}{5}$

To find : The value of $\frac{4\tan\theta-5\cos\theta}{\sec\theta+4\cot\theta}$

Solution :

$\sin\theta=\frac{4}{5}$

According to trigonometric properties,

$\sin\theta=\frac{4}{5}=\frac{P}{H}$

i.e Perpendicular P=4, Hypotenuse H=5

Apply Pythagoras theorem,

$B=\sqrt{H^2-P^2}$

$B=\sqrt{5^2-4^2}$

$B=\sqrt{25-16}$

$B=\sqrt{9}$

$B=3$

We know,

$\cos\theta=\frac{B}{H}=\frac{3}{5}$

$\tan\theta=\frac{P}{B}=\frac{4}{3}$

$\cot\theta=\frac{B}{P}=\frac{3}{4}$

$\sec\theta=\frac{H}{B}=\frac{5}{3}$

Substitute the value in the expression,

$=\frac{4(\frac{4}{3})-5(\frac{3}{5})}{(\frac{5}{3})+4(\frac{3}{4})}$

$=\frac{\frac{16}{3}-3}{\frac{5}{3}+3}$

$=\frac{\frac{16-9}{3}}{\frac{5+9}{3}}$

$=\frac{\frac{7}{3}}{\frac{14}{3}}$

$=\frac{7}{3}\times \frac{3}{14}$

$=\frac{1}{2}$

Therefore, $\frac{4\tan\theta-5\cos\theta}{\sec\theta+4\cot\theta}=\frac{1}{2}$