Question

do this short clearly process I will tick as branilist answer

Answer 1

Answer:

i am unable to understand your writing

so i solve that i understand

Answer 2

We only have,

$\tan\theta=\dfrac{1}{\sqrt7}$

where θ is an acute angle.

We take the reciprocal to get cotθ.

$\cot\theta=\sqrt7$

And we take it's square.

$\cot^2\theta=7$

Also, we remember,

$\cot\theta=\dfrac{\csc\theta}{\sec\theta}\ \ \ \ \ \Longrightarrow\ \ \ \ \ \cot^2\theta=\dfrac{\csc^2\theta}{\sec^2\theta}$

Now,

$\begin{aligned}&\frac{\csc^2\theta-\sec^2\theta}{\csc^2\theta+\sec^2\theta}\\ \\ \Longrightarrow\ \ &\frac{\frac{\csc^2\theta-\sec^2\theta}{\sec^2\theta}}{\frac{\csc^2\theta+\sec^2\theta}{\sec^2\theta}}\\ \\ \Longrightarrow\ \ &\frac{\frac{\csc^2\theta}{\sec^2\theta}-\frac{\sec^2\theta}{\sec^2\theta}}{\frac{\csc^2\theta}{\sec^2\theta}+\frac{\sec^2\theta}{\sec^2\theta}}\end{aligned}$

$\begin{aligned}\Longrightarrow\ \ &\frac{\cot^2\theta-1}{\cot^2\theta+1}\\ \\ \Longrightarrow\ \ &\frac{7-1}{7+1}\\ \\ \Longrightarrow\ \ &\frac{6}{8}\\ \\ \Longrightarrow\ \ &\large\text{ \bold{\frac{3}{4}} }\end{aligned}$

Hence 3/4 is the answer.