Question

if 10 theta = 1/ root 7
show that cosec ² theta - sec ²theta
upon cosec²theta + sec²theta =3/4​

Answer 1

♣ 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

________________________________________

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 !!!

Answer 2

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 !!!