Question

x = a tan^2 alpha + b sec^2 alpha
then x+a/x-b = ​

Answer 1

Answer:

$x + a = a + a \: tan {}^{2} \alpha + b \: sec {}^{2} \alpha = a(1 + {tan}^{2} \alpha ) + b \: sec {}^{2} \alpha = a \: sec {}^{2} \alpha + b \: sec {}^{2} \alpha = (a + b) \: sec {}^{2} \alpha$

Now,

$x - b = a \: {tan}^{2} \alpha + b \: {sec}^{2} \alpha - b = a \: {tan}^{2} \alpha + b( {sec}^{2} \alpha - 1) = a \: {tan}^{2} \alpha + b \: {tan}^{2} \alpha = (a + b) \: {tan}^{2} \alpha$

so,

$\frac{x + a}{x - b} = \frac{(a + b) {sec}^{2} \alpha }{(a + b) {tan}^{2} \alpha } = \frac{1}{ {cos}^{2} \alpha } \times \frac{ {cos}^{2} \alpha }{ {sin}^{2} \alpha } = {cosec}^{2} \alpha \: \: ans.$

2nd option correct.