Question

hey
pls help me solve this question
thank you ​

Answer 1

d, is the answer

if my answer is correct me

please Mark as brainlist

Answer 2

Answer:

The answer is $\frac{1}{2}$

Explanation:

Given $\tan\theta=\frac{1}{\sqrt 3}$

So,

$\frac{\cos ec^2\theta-\sec^2\theta }{\cos ec^2\theta-\sec^2\theta }=\frac{(1+\cot^2\theta)-(1+\tan^2\theta)}{(1+\cot^2\theta)+(1+\tan^2\theta)}$  

( using $\sec^2\theta-\cot^2\theta=1$ and $\cos ec^2\theta-\cot^2\theta=1$ )

$=\frac{\cot^2\theta-\tan^2\theta}{2+\cot^2\theta+\tan^2\theta}\\=\frac{\frac{1}{\tan^2\theta}-\tan^2\theta}{2+\frac{1}{\tan^2\theta}+\tan^2\theta}\\=\frac{3-\frac{1}{3}}{2+3+\frac{1}{3}}=\frac{1}{2}$