Question

if 15 tan theta =14.Evaluate 2sintheta cos theta/cos square theta-sin theta. please explain with steps this is an emergency!!!!!!!!​

Answer 1

Here, $15\tan \theta = 14$

This implies,

$\tan \theta =\frac{14}{15}=\frac{Perpedicular}{Base}$

Assume x be the hypotenuse.

Then, by the Pythagoras theorem,

$x=\sqrt{14^2+15^2}$

  $=\sqrt{196+225}$

  $=\sqrt{421}$

Thus,

$\sin \theta =\frac{Perpedicular}{Hypotenuse}=\frac{14}{\sqrt{421}}$

$\cos \theta =\frac{Base}{Hypotenuse}=\frac{15}{\sqrt{421}}$

So,

$\frac{2\sin \theta \cos \theta}{\cos^2 \theta -\sin \theta}$

$=\frac{2\times \frac{14}{\sqrt{421}}\times \frac{15}{\sqrt{421}}}{\frac{225}{421}-\frac{14}{\sqrt{421}}}$

$=\frac{420}{225-14\sqrt{421}}$

Therefore, $\frac{2\sin \theta \cos \theta}{\cos^2 \theta -\sin \theta}= \frac{420}{225-14\sqrt{421}}$

Answer 2

Given:

15 tan theta = 14.

To find:

Evaluate 2sintheta cos theta/cos square theta-sin theta.

Solution:

From given, we have,

15 tan theta = 14

tan θ = 14/15

we use the trigonometric property,

tan θ = opposite side/adjacent side

hypotenuse = √(opposite side² + adjacent side²) = √(14² + 15²) = √421

sin θ = O/H = 14/√421

cos θ = A/H = 15/√421

Given, 2sintheta cos theta/cos square theta-sin theta.

⇒ 2 sin θ cos θ / cos² θ-sin θ

$= \dfrac{2\cdot \frac{14}{\sqrt{421}}\cdot \frac{15}{\sqrt{421}}}{\left(\frac{15}{\sqrt{421}}\right)^2-\frac{14}{\sqrt{421}}}\\\\= \dfrac{2\cdot \frac{14}{\sqrt{421}}\cdot \frac{15}{\sqrt{421}}}{(\frac{225}{{421}})-\frac{14}{\sqrt{421}}}\\\\= \dfrac{\frac{420}{421}}{\frac{225-14\sqrt{421}}{421}}\\\\\\=\dfrac{420}{225-14\sqrt{421}}$

$\therefore \dfrac{2sin\theta cos \theta }{cos^2 \theta -sin \theta}=\dfrac{420}{225-14\sqrt{421}}$