Question

If $\tan \theta =\dfrac{a}{b}$ , then $\dfrac{a\sin \theta +b\cos \theta }{a\sin \theta -b\cos \theta }$ is equal to

a)$\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}$
b)$\dfrac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}}$
c)$\dfrac{a+b}{a-b}$
d)$\dfrac{a-b}{a+b}$

Answer 1

Hint: Here, first we have to divide the numerator and denominator of the function $\dfrac{a\sin \theta +b\cos \theta }{a\sin \theta -b\cos \theta }$ by $\cos \theta $ and also use the concept that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$. Then, by simplification we can obtain the value of $\dfrac{a\sin \theta +b\cos \theta }{a\sin \theta -b\cos \theta }$.

Complete step-by-step answer:
Here, we are given that $\tan \theta =\dfrac{a}{b}$.

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

Here, divide the numerator and denominator of the function $\dfrac{a\sin \theta +b\cos \theta }{a\sin \theta -b\cos \theta }$ by $\cos \theta $, we obtain:

$\dfrac{a\sin \theta +b\cos \theta }{a\sin \theta -b\cos \theta }=\dfrac{\dfrac{a\sin \theta

+b\cos \theta }{\cos \theta }}{\dfrac{a\sin \theta -b\cos \theta }{\cos \theta }}$

Next, by splitting the function,

$\Rightarrow \dfrac{a\sin \theta +b\cos \theta }{a\sin \theta -b\cos \theta

}=\dfrac{\dfrac{a\sin \theta }{\cos \theta }+\dfrac{b\cos \theta }{\cos \theta }}{\dfrac{a\sin

\theta }{\cos \theta }-\dfrac{b\cos \theta }{\cos \theta }}$

We know that,

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

Now, by substituting this in the above function,

$\Rightarrow \dfrac{a\sin \theta +b\cos \theta }{a\sin \theta -b\cos \theta }=\dfrac{a\tan \theta +\dfrac{b\cos \theta }{\cos \theta }}{a\tan \theta -\dfrac{b\cos \theta }{\cos \theta }}$
Next, by cancellation of $\cos \theta $,

$\Rightarrow \dfrac{a\sin \theta +b\cos \theta }{a\sin \theta -b\cos \theta }=\dfrac{a\tan \theta +b}{a\tan \theta -b}$

Now, by substituting the value of $\tan \theta =\dfrac{a}{b}$,

$\begin{align}

  & \Rightarrow \dfrac{a\sin \theta +b\cos \theta }{a\sin \theta -b\cos \theta

}=\dfrac{a\times \dfrac{a}{b}+b}{a\times \dfrac{a}{b}-b} \\

 & \Rightarrow \dfrac{a\sin \theta +b\cos \theta }{a\sin \theta -b\cos \theta

}=\dfrac{\dfrac{{{a}^{2}}}{b}+b}{\dfrac{{{a}^{2}}}{b}-b} \\

\end{align}$

Next, by taking LCM,

$\Rightarrow \dfrac{a\sin \theta +b\cos \theta }{a\sin \theta -b\cos \theta

}=\dfrac{\dfrac{{{a}^{2}}+{{b}^{2}}}{b}}{\dfrac{{{a}^{2}}-{{b}^{2}}}{b}}$

We know that,

$\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}$

Hence, we can write:

$\dfrac{a\sin \theta +b\cos \theta }{a\sin \theta -b\cos \theta

}=\dfrac{{{a}^{2}}+{{b}^{2}}}{b}\times \dfrac{b}{{{a}^{2}}-{{b}^{2}}}$

Next, by cancellation,

$\Rightarrow \dfrac{a\sin \theta +b\cos \theta }{a\sin \theta -b\cos \theta

}=\dfrac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}}$

Therefore, we can say that the value of $\dfrac{a\sin \theta +b\cos \theta }{a\sin \theta

-b\cos \theta }=\dfrac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}}$.

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



Note: Here, we can also solve this problem by using the definition of $\tan \theta $ that is, $\tan \theta =\dfrac{opposite\text{ }side}{adjacent\text{ }side}$. First we have to find the hypotenuse of the triangle by using the concept of Pythagoras theorem, where the square of the hypotenuse is the sum of the squares of adjacent side and opposite side. Then, we have to find the values of $\sin \theta $ and $\cos \theta $ and substitute in the function $\dfrac{a\sin \theta +b\cos \theta }{a\sin \theta -b\cos \theta }$.