Answer :
tan theta=1/root7=P/B
so, using pythagorean H=2root2.
putting the values we will have,
8/7*8*7/48
=4/3
♣ Qᴜᴇꜱᴛɪᴏɴ :
- If tan θ = [tex]\sf{\dfrac{1}{\sqrt{7}}}[/tex], Show that [tex]\sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }=\dfrac{3}{4}}[/tex]
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♣ ᴀɴꜱᴡᴇʀ :
We know :
[tex]\large\boxed{\sf{tan\theta=\dfrac{Height}{Base}}}[/tex]
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 : [tex]\sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }=\dfrac{3}{4}}[/tex]
Also :
[tex]\large\boxed{\sf{cosec\theta=\dfrac{Hypotenuse}{Height}}}[/tex]
[tex]\large\boxed{\sf{sec\theta=\dfrac{Hypotenuse}{Base}}}[/tex]
From this we get :
[tex]\large\boxed{\sf{cosec^2\theta=\left(\dfrac{Hypotenuse}{Height}\right)^2}}[/tex]
[tex]\large\boxed{\sf{sec^2\theta=\left(\dfrac{Hypotenuse}{Base}\right)^2}}[/tex]
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
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First cosec²θ :
[tex]\large\boxed{\sf{cosec^2\theta=\left(\dfrac{Hypotenuse}{Height}\right)^2}}[/tex]
[tex]\sf{cosec^2\theta=\left(\dfrac{\sqrt{8}}{1}\right)^2}[/tex]
[tex]\sf{cosec^2\theta=\dfrac{8}{1}}[/tex]
cosec²θ = 8
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Now sec²θ :
[tex]\large\boxed{\sf{sec^2\theta=\left(\dfrac{Hypotenuse}{Base}\right)^2}}[/tex]
[tex]\sf{sec^2\theta=\left(\dfrac{\sqrt{8}}{\sqrt{7}}\right)^2}[/tex]
[tex]\sf{sec^2\theta=\dfrac{8}{7}}[/tex]
sec²θ = 8/7
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Now Proving :
[tex]\sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }=\dfrac{3}{4}}[/tex]
Taking L.H.S :
[tex]\sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }}[/tex]
[tex]=\sf{\dfrac{8 - sec ^2\theta}{8 + sec^2\theta }}[/tex]
[tex]=\sf{\dfrac{8 - \dfrac{8}{7}}{8 + \dfrac{8}{7} }}[/tex]
[tex]=\sf{\dfrac{\dfrac{48}{7}}{\dfrac{64}{7} }}[/tex]
[tex]\sf{=\dfrac{48\times \:7}{7\times \:64}}[/tex]
[tex]\sf{=\dfrac{48}{64}}[/tex]
[tex]\bf{=\dfrac{3}{4}}[/tex]