Math, asked by omkar6290, 8 months ago

If tan theta = 1/root 7 show that cosec^2 theta - sec^2 theta/cosec^2 theta+sec^2 theta equals to 3/4

Answers

Answered by hd273616
0

67o7787Harshdeep

Sutton in my opinion as

Answered by Anonymous
112

♣ Qᴜᴇꜱᴛɪᴏɴ :

  • If tan θ  = \sf{\dfrac{1}{\sqrt{7}}}, Show that \sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }=\dfrac{3}{4}}

★═════════════════★

♣ ᴀɴꜱᴡᴇʀ :

We know :

\large\boxed{\sf{tan\theta=\dfrac{Height}{Base}}}

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}}

Also :

\large\boxed{\sf{cosec\theta=\dfrac{Hypotenuse}{Height}}}

\large\boxed{\sf{sec\theta=\dfrac{Hypotenuse}{Base}}}

From this we get :

\large\boxed{\sf{cosec^2\theta=\left(\dfrac{Hypotenuse}{Height}\right)^2}}

\large\boxed{\sf{sec^2\theta=\left(\dfrac{Hypotenuse}{Base}\right)^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}}

\sf{cosec^2\theta=\left(\dfrac{\sqrt{8}}{1}\right)^2}

\sf{cosec^2\theta=\dfrac{8}{1}}

cosec²θ = 8

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Now sec²θ :

\large\boxed{\sf{sec^2\theta=\left(\dfrac{Hypotenuse}{Base}\right)^2}}

\sf{sec^2\theta=\left(\dfrac{\sqrt{8}}{\sqrt{7}}\right)^2}

\sf{sec^2\theta=\dfrac{8}{7}}

sec²θ = 8/7

________________________________________

Now Proving :

\sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }=\dfrac{3}{4}}

Taking L.H.S :

\sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }}

=\sf{\dfrac{8 - sec ^2\theta}{8 + sec^2\theta }}

=\sf{\dfrac{8 - \dfrac{8}{7}}{8 + \dfrac{8}{7} }}

=\sf{\dfrac{\dfrac{48}{7}}{\dfrac{64}{7} }}

\sf{=\dfrac{48\times \:7}{7\times \:64}}

\sf{=\dfrac{48}{64}}

\bf{=\dfrac{3}{4}}

= R.H.S

Hence Proved !!!

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