Question

If $tan\Theta=\frac{1}{\sqrt{7} }$, show that $\frac{cosec^{2}\Theta-sec^{2}\Theta }{cosec^{2}\Theta+sec^{2}\Theta } =\frac{3}{4}$.

Answer 1

SOLUTION :  

Given: tan θ = 1/√7

tan θ = 1/√7 = P/ B  = BC/AB

Draw a right ∆ABC, ∠B = 90°

Base(AB) = √7 unit & Perpendicular(BC) = 1 unit

In ∆ABC,  

AC² = AB² + BC²

[By using Pythagoras theorem]

AC² = (√7)² + (1)²

AC² = 7 + 1

AC² = 8

AC =√8

AC = 2√2

Now, cosec θ = H/P = AC/BC = 2√2 /1

sec θ = H /B = AC/ AB = 2√2 /√7

L.H.S = cosec²θ - sec²θ / cosec²θ +sec² θ  

= (2√2)² - (2√2 /√7)² / (2√2)² + (2√2 /√7)²

= (4×2) - (4×2/ 7) /  (4×2) +  (4×2/ 7)

=( 8 - 8/7 ) / (8 + 8/7)

=[(56 - 8) /7] / [(56 + 8) /7]  

= (48 /7) / (64/7)

= 48/64

= 3 / 4  

= R.H.S  

Hence, cosec²θ - sec²θ / cosec²θ +sec² θ = ¾

HOPE THIS ANSWER WILL HELP YOU….