Question

how to solve this problem ​

Answer 1

(instead of θ, I use A,.)

Answer:

$\frac{3}{4}$

Step-by-step explanation:

Given :

$tan A = \frac{1}{\sqrt{7}}$

Then, the value of :

$\frac{cosec^2A - sec^2A}{cosec^2A + sec^2A}$

Solution :

We know that,

$tanA= \frac{sinA}{cosA} = \frac{1}{\sqrt{7} }$

⇒ sin A = 1 × k (for some constant k)

⇒ cos A = √7 × k = k√7

⇒ $sinA = \frac{1}{cosecA} = \frac{1}{k}$

⇒ $cosA = \frac{1}{secA} = \frac{1}{k\sqrt{7}}$

So,.

$\frac{cosec^2A - sec^2A}{cosec^2A + sec^2A}$ = $\frac{(\frac{1}{k}) ^2 - (\frac{1}{k\sqrt{7} } )^2}{(\frac{1}{k}) ^2 + (\frac{1}{k\sqrt{7} } )^2}$

⇒ $\frac{\frac{1}{k^2} - \frac{1}{7k^2} }{\frac{1}{k^2} + \frac{1}{7k^2}}$

⇒ $\frac{\frac{7}{7k^2} - \frac{1}{7k^2} }{\frac{7}{7k^2} + \frac{1}{7k^2}}$

⇒ $\frac{\frac{6}{7k^2}}{\frac{8}{7k^2} }$

⇒ $\frac{6}{7k^2} \div \frac{8}{7k^2}$

⇒ $\frac{6}{7k^2} \times \frac{7k^2}{8}$

⇒ $\frac{6}{8}=\frac{3}{4}$