Answer:
㋡
Check Answer
♣ Qᴜᴇꜱᴛɪᴏɴ :
If tan θ = \sf{\dfrac{1}{\sqrt{7}}} 
7
 
 
1
 
 , Show that \sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }=\dfrac{3}{4}} 
cosec 
2
 θ+sec 
2
 θ
cosec 
2
 θ−sec 
2
 θ
 
 = 
4
3
 
 
★═════════════════★
♣ ᴀɴꜱᴡᴇʀ :
We know :
\large\boxed{\sf{tan\theta=\dfrac{Height}{Base}}} 
tanθ= 
Base
Height
 
 
 
 
So comparing this formula and value of tan θ from question, we get :
Height = 1
Base = √7
Now we need to Prove the value of : \sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }=\dfrac{3}{4}} 
cosec 
2
 θ+sec 
2
 θ
cosec 
2
 θ−sec 
2
 θ
 
 = 
4
3
 
 
Also :
\large\boxed{\sf{cosec\theta=\dfrac{Hypotenuse}{Height}}} 
cosecθ= 
Height
Hypotenuse
 
 
 
 
\large\boxed{\sf{sec\theta=\dfrac{Hypotenuse}{Base}}} 
secθ= 
Base
Hypotenuse
 
 
 
 
From this we get :
\large\boxed{\sf{cosec^2\theta=\left(\dfrac{Hypotenuse}{Height}\right)^2}} 
cosec 
2
 θ=( 
Height
Hypotenuse
 
 ) 
2
 
 
 
\large\boxed{\sf{sec^2\theta=\left(\dfrac{Hypotenuse}{Base}\right)^2}} 
sec 
2
 θ=( 
Base
Hypotenuse
 
 ) 
2
 
 
 
But we have Height and Base, we dont have Hypotenuse.
Hypotenuse can be found by using Pythagoras Theorem
Pythagoras Theorem states that :
Hypotenuse² = Side² + Side²
For our question :
Hypotenuse² = Height² + Base²
Hypotenuse² = 1² + √7²
Hypotenuse² = 1 + 7
Hypotenuse² = 8
√Hypotenuse² = √8
Hypotenuse = √8
➢ Let's find value's of cosec²θ and sec²θ
________________________________________
First cosec²θ :
\large\boxed{\sf{cosec^2\theta=\left(\dfrac{Hypotenuse}{Height}\right)^2}} 
cosec 
2
 θ=( 
Height
Hypotenuse
 
 ) 
2
 
 
 
\sf{cosec^2\theta=\left(\dfrac{\sqrt{8}}{1}\right)^2}cosec 
2
 θ=( 
1
8
 
 
 
 ) 
2
 
\sf{cosec^2\theta=\dfrac{8}{1}}cosec 
2
 θ= 
1
8
 
 
cosec²θ = 8
________________________________________
Now sec²θ :
\large\boxed{\sf{sec^2\theta=\left(\dfrac{Hypotenuse}{Base}\right)^2}} 
sec 
2
 θ=( 
Base
Hypotenuse
 
 ) 
2
 
 
 
\sf{sec^2\theta=\left(\dfrac{\sqrt{8}}{\sqrt{7}}\right)^2}sec 
2
 θ=( 
7
 
 
8
 
 
 
 ) 
2
 
\sf{sec^2\theta=\dfrac{8}{7}}sec 
2
 θ= 
7
8
 
 
sec²θ = 8/7
________________________________________
Now Proving :
\sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }=\dfrac{3}{4}} 
cosec 
2
 θ+sec 
2
 θ
cosec 
2
 θ−sec 
2
 θ
 
 = 
4
3
 
 
Taking L.H.S :
\sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }} 
cosec 
2
 θ+sec 
2
 θ
cosec 
2
 θ−sec 
2
 θ
 
 
=\sf{\dfrac{8 - sec ^2\theta}{8 + sec^2\theta }}= 
8+sec 
2
 θ
8−sec 
2
 θ
 
 
=\sf{\dfrac{8 - \dfrac{8}{7}}{8 + \dfrac{8}{7} }}= 
8+ 
7
8
 
 
8− 
7
8
 
 
 
 
=\sf{\dfrac{\dfrac{48}{7}}{\dfrac{64}{7} }}= 
7
64
 
 
7
48
 
 
 
 
\sf{=\dfrac{48\times \:7}{7\times \:64}}= 
7×64
48×7
 
 
\sf{=\dfrac{48}{64}}= 
64
48
 
 
\bf{=\dfrac{3}{4}}= 
4
3
 
 
= R.H.S
Hence Proved !!!