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 !!!