Question

If $5\tan \theta -4=0$ then the value of $\dfrac{5\sin \theta -4\cos \theta }{5\sin \theta +4\cos \theta }$ is equal to,

a)$\dfrac{5}{3}$
b)$\dfrac{5}{6}$
c)0
d)$\dfrac{1}{6}$

Answer 1

Hint: Here, first we have to find the value of $\tan \theta $ from $5\tan \theta -4=0$. Then, divide the numerator and denominator of the function $\dfrac{5\sin \theta -4\cos \theta }{5\sin \theta +4\cos \theta }$ by $\cos \theta $. Next, we have apply the formula $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and also substitute the value of $\tan \theta $ that we obtained from $5\tan \theta -4=0$. Then, do the simplification to obtain the answer.

Complete step-by-step answer:
Here, we are given that $5\tan \theta -4=0$.

Now, we have to find the value of $\dfrac{5\sin \theta -4\cos \theta }{5\sin \theta +4\cos \theta }$.

First consider,

$5\tan \theta -4=0$

Now, by taking -4 to the right side -4 becomes 4,

$\Rightarrow 5\tan \theta =4$

Next, by cross multiplication,

$\Rightarrow \tan \theta =\dfrac{4}{5}$

The corresponding figure is as follows:

Next, consider the function, $\dfrac{5\sin \theta -4\cos \theta }{5\sin \theta +4\cos \theta }$.

Now, divide the numerator and denominator of the function by $\cos \theta $, we obtain:$\dfrac{5\sin \theta -4\cos \theta }{5\sin \theta +4\cos \theta }=\dfrac{5\dfrac{\sin \theta }{\cos \theta }-4\dfrac{\cos \theta }{\cos \theta }}{5\dfrac{\sin \theta }{\cos \theta }+4\dfrac{\cos \theta }{\cos \theta }}$

We know that,

$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$

Now, by substituting the above formula, we get:

$\dfrac{5\sin \theta -4\cos \theta }{5\sin \theta +4\cos \theta }=\dfrac{5\tan \theta -4\dfrac{\cos \theta }{\cos \theta }}{5\tan \theta +4\dfrac{\cos \theta }{\cos \theta }}$

Next, by cancellation,

$\Rightarrow \dfrac{5\sin \theta -4\cos \theta }{5\sin \theta +4\cos \theta }=\dfrac{5\tan \theta -4}{5\tan \theta +4}$

We got that $\tan \theta =\dfrac{4}{5}$, hence, by substituting this value we obtain:

$\Rightarrow \dfrac{5\sin \theta -4\cos \theta }{5\sin \theta +4\cos \theta }=\dfrac{5\times \dfrac{4}{5}-4}{5\times \dfrac{4}{5}+4}$

Now, by cancellation,

$\begin{align}

  & \Rightarrow \dfrac{5\sin \theta -4\cos \theta }{5\sin \theta +4\cos \theta }=\dfrac{4-4}{4+4} \\

 & \Rightarrow \dfrac{5\sin \theta -4\cos \theta }{5\sin \theta +4\cos \theta }=\dfrac{0}{8} \\

 & \Rightarrow \dfrac{5\sin \theta -4\cos \theta }{5\sin \theta +4\cos \theta }=0 \\

\end{align}$

Therefore, we can say that the value of $\dfrac{5\sin \theta -4\cos \theta }{5\sin \theta +4\cos \theta }$ when $5\tan \theta -4=0$ is 0.

Hence, the correct answer for this question is option (c).


Note: Here, we can also solve this question by applying the formula for $\tan \theta =\dfrac{Opposite\text{ }side}{Adjacent\text{ }side}$. Then, find the hypotenuse by Pythagoras theorem, where the square of the hypotenuse is the sum of the squares of the opposite side and adjacent side. Next, find the values of $\sin \theta $ and $\cos \theta .$ Substitute the values in the function $\dfrac{5\sin \theta -4\cos \theta }{5\sin \theta +4\cos \theta }$ to obtain the answer.