Question 141

If $$4\tan\theta=3,\frac{5\sin\theta-3\cos\theta}{5\sin\theta+3\cos\theta}$$ is equal to:

Solution

$$4\tan\theta\ =3.$$

or,$$\frac{\sin\ \theta\ \ }{\cos\ \theta\ }=\ \frac{\ 3}{4}.$$

Let say,$$\sin\theta\ =3k\ and\ \cos\theta\ =4k.$$

So,$$\ \frac{\ 5\sin\ \theta\ -3\cos\ \theta\ }{5\sin\ \theta\ +3\cos\ \theta\ }$$

$$\ =\frac{\ 5\times\ 3k\ -3\times\ 4k\ }{5\times\ 3k\ +3\times\ 4k\ }$$

$$\ =\frac{\ 15k\ -12k\ }{15k\ +12k\ }$$

$$\ =\frac{\ 3k\ }{27k\ }$$

$$\ =\frac{\ 1\ }{9}.$$

A is correct choice.

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