Question

please give the brief answer for the following​

Answer 1

Answer:

181/109

Step-by-step explanation:

Now

$sin\theta < 0$

Thus, Either θ in 3rd quadrant or in 4th quadrant

According to question θ is not in 3rd Quadrant

So it must be in 4th quadrant

Now for θ in fourth quadrant

cosθ > 0

tanθ < 0

sinθ < 0

$cos\theta = \pm\sqrt{1 - sin^{2}\thata}\\\\cos\theta = \pm\sqrt{1 - (\frac{-5}{13})^{2}}\\\\cos\theta = \pm\sqrt{1 - \frac{25}{169}}\\\\cos\theta = \pm\sqrt{\frac{144}{169}\\}\\\\cos\theta = \pm\frac{12}{13}$

we are also given that cosθ > 0

$cos\theta = \frac{12}{13}$

so $tan\theta = \frac{sin\theta}{cos\theta}$

$tan\theta = \frac{\frac{-5}{13}}{\frac{12}{13}}\\\\tan\theta = \frac{-5}{12}$

Now note some identities

$tan(90 + \theta) = -cot\theta = \frac{-1}{tan\theta}\\\\sin(180-\theta) = sin\theta\\\\sin(270 - \theta) = -cos\theta\\\\cosec(360 - \theta) = \frac{-1}{sin\theta}$

$\frac{tan(90+\theta) - sin(180-\theta)}{sin(270-\theta)+cosec(360-\theta)}}\\\\\frac{\frac{-1}{tan\theta} - sin\theta}{-cos\theta - \frac{1}{sin\theta}}$

Now put values

$\frac{\frac{12}{5}+\frac{5}{13}}{-\frac{12}{13} + \frac{13}{5}}\\\\\frac{\frac{181}{13\times5}}{\frac{109}{13\times5}}\\\\\frac{181}{109}$

so 181/109 is correct answer!