Question

If we have an expression $5\tan \theta = 4$, then $\dfrac{{5\sin \theta - 3\cos \theta }}{{5\sin \theta + 2\cos \theta }} =$
$A)0$
$B)1$
$C)\dfrac{1}{6}$
$D)6$

Answer 1

First, we need to analyze the given information which is in the trigonometric form.
The trigonometric functions are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
We can simply find the value of the tangent first, and then convert the required equation to tangent then substitute the values to get the results required.
Formula used:
$\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta$

Complete step-by-step solution:
Since from the given that we have, $5\tan \theta = 4$
Just divide both the side using the number $5$ then we get $\dfrac{5}{5}\tan \theta = \dfrac{4}{5}$
Now by the division operation, $\dfrac{5}{5} = 1$ then we get $\tan \theta = \dfrac{4}{5}$ which is the value of the tangent.
Now we need to find the value of $\dfrac{{5\sin \theta - 3\cos \theta }}{{5\sin \theta + 2\cos \theta }}$
Take the numerator value $5\sin \theta - 3\cos \theta$ and divide with the trigonometric function cosine, then we get $\dfrac{{5\sin \theta }}{{\cos \theta }} - \dfrac{{3\cos \theta }}{{\cos \theta }}$ again by the make use of the division operation that same values in both numerator and denominator will get cancel thus we get $\dfrac{{5\sin \theta }}{{\cos \theta }} - \dfrac{{3\cos \theta }}{{\cos \theta }} \Rightarrow \dfrac{{5\sin \theta }}{{\cos \theta }} - 3$
Since by the trigonometric formulas we have, $\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta$ and substituting these values we get $\dfrac{{5\sin \theta }}{{\cos \theta }} - 3 \Rightarrow 5\tan \theta - 3$ which is the simplified numerator value.
Similarly, tale the denominator value $5\sin \theta + 2\cos \theta$ and divide with the trigonometric function cosine, then we get $\dfrac{{5\sin \theta }}{{\cos \theta }} + \dfrac{{2\cos \theta }}{{\cos \theta }}$ again by the make use of the division operation that same values in both numerator and denominator will get cancel thus we get $\dfrac{{5\sin \theta }}{{\cos \theta }} + \dfrac{{2\cos \theta }}{{\cos \theta }} \Rightarrow \dfrac{{5\sin \theta }}{{\cos \theta }} + 2$
Since by the trigonometric formulas we have, $\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta$ and substituting these values we get $\dfrac{{5\sin \theta }}{{\cos \theta }} + 2 \Rightarrow 5\tan \theta + 2$ which is the simplified numerator value.
Therefore, we have $\dfrac{{5\sin \theta - 3\cos \theta }}{{5\sin \theta + 2\cos \theta }} = \dfrac{{5\tan \theta - 3}}{{5\tan \theta + 2}}$
Since from the given, we have the value of tangent as $\tan \theta = \dfrac{4}{5}$ and substituting this value we get $\dfrac{{5\tan \theta - 3}}{{5\tan \theta + 2}} = \dfrac{{5(\dfrac{4}{5}) - 3}}{{5(\dfrac{4}{5}) + 2}}$ and canceling the common terms we get $\dfrac{{5\tan \theta - 3}}{{5\tan \theta + 2}} = \dfrac{{5(\dfrac{4}{5}) - 3}}{{5(\dfrac{4}{5}) + 2}} \Rightarrow \dfrac{{4 - 3}}{{4 + 2}} = \dfrac{1}{6}$
Thus, we get $\dfrac{{5\sin \theta - 3\cos \theta }}{{5\sin \theta + 2\cos \theta }} = \dfrac{1}{6}$
Therefore, the option $C)\dfrac{1}{6}$ is correct.

Note: In total there are six trigonometric values which are sine, cos, tan, sec, cosec, cot while all the values have been relation like $\dfrac{{\sin }}{{\cos }} = \tan$and $\tan = \dfrac{1}{{\cot }}$
Also, the inverse functions of the trigonometric can be represented as $\tan \theta = \dfrac{4}{5} \Rightarrow \theta = {\tan ^{ - 1}}\dfrac{4}{5}$
The other trigonometric functions as $\dfrac{1}{{\sin }} = \cos ec,\dfrac{1}{{\cos }} = \sec$