Question

If 7 tan theta is equal to 4 find the value of 7 sin theta minus 3 cos theta by 7 sin theta plus 3 cos theta

Answer 1

Step-by-step explanation:

hope this will help you.

Answer 2

Given:

7tanθ=4

To find:

(7sinθ-3cosθ) / (7sinθ+3cosθ)

Solution:

We know that,

tanθ=$\frac{perpendicular}{base}$

We have,

7tanθ=4

tanθ$=\frac{4}{7}$

So, we can say that perpendicular is 4units and base is 7units.

Now,

$hypotenuse=\sqrt{(perpendicular)^2+(base)^2}$

$=\sqrt{4^2+7^2}$

$=\sqrt{16+49}$

$=\sqrt{65}$units

We know that,

sinθ$=\frac{perpendicular}{hypotenuse}$

=$\frac{4}{\sqrt{65} }$

And we know,

cosθ$=\frac{base}{hypotenuse}$

=$\frac{7}{\sqrt{65} }$

Now evaluating the numerator,

7sinθ-3cosθ

=7×$\frac{4}{\sqrt{65} }$-4×$\frac{7}{\sqrt{65} }$

=$\frac{28-21}{\sqrt{65} }$

$=\frac{7}{\sqrt{65} }$

Now, evaluating the denominator,

7sinθ+3cosθ

=7×$\frac{4}{\sqrt{65} }$+4×$\frac{7}{\sqrt{65} }$

=$\frac{28+21}{\sqrt{65} }$

$=\frac{49}{\sqrt{65} }$

So,

(7sinθ-3cosθ) / (7sinθ+3cosθ)

=$\frac{\frac{7}{\sqrt{65} } }{\frac{49}{\sqrt{65} } }$

=$\frac{7}{\sqrt{65} }$×$\frac{\sqrt{65} }{49}$

$=\frac{1}{7}$

Hence, the required value of (7sinθ-3cosθ) / (7sinθ+3cosθ) is $\frac{1}{7}$.