Question

if sec theta =2 / root 3 ,then find the value of 1- cos theta / 1+ cos theta ​

Answer 1

S O L U T I O N :

Given :

sec Ѳ = 2/√3

Explanation :

According to the question attachment a figured of right angled triangle.

As we know that sec Ѳ = Hypotenuse/base = AC/AB

sec Ѳ = 2/√3 = AC/AB

Using by Pythagoras Theorem :

➝ (Hypotenuse)² = (Base)² + (Perpendicular)²

➝ (AC)² = (AB)² + (BC)²

➝ (2)² = (√3)² + (BC)²

➝ 4 = √3 × √3 + (BC)²

➝ 4 = 3 + (BC)²

➝ (BC)² = 4 - 3

➝ (BC)² = 1

➝ BC = √1

Now,

➝ 1 - cos Ѳ/1 + cos Ѳ

➝ 1 - AB/AC/1 + AB/AC

➝ 1 - √3/2/1 + √3/2

➝ 2-√3/2/2 + √3/2

➝ 2 - √3/2 × 2/2 + √3

➝ 2 - √3/2 + √3

• Rationalisation :

➝ 2 - √3/2 + √3 × 2- √3/2 - √3

➝ (2 - √3)²/(2)² - (√3)²

➝ (2)² + (√3)² - 2 × 2 × √3/4 - 3

➝ 4 + 3 - 4√3 / 1

➝ 7 - 4√3/1

➝ 7 - 4√3

Thus,

The value of 1 - cos Ѳ/1 + cos Ѳ will be 7 - 4√3 .

Answer 2

${\large{\bold{\sf{\underline{Question}}}}}$

Sec θ = ${\bold{\sf{\dfrac{2}{\sqrt{3}}}}}$ then we have to find the value of ${\bold{\sf{\dfrac{1-cos \theta}{1+cos \theta}}}}$

${\large{\bold{\sf{\underline{Given \; that}}}}}$

Sec θ = ${\bold{\sf{\dfrac{2}{\sqrt{3}}}}}$

${\large{\bold{\sf{\underline{To \; find}}}}}$

${\bold{\sf{\dfrac{1-cos \theta}{1+cos \theta}}}}$

${\large{\bold{\sf{\underline{Solution}}}}}$

${\bold{\sf{\dfrac{1-cos \theta}{1+cos \theta}}}}$ = 7 - 4√3

${\large{\bold{\sf{\underline{Full \; Solution}}}}}$

We know that sec θ = ${\bold{\sf{\dfrac{Hypotenuse}{Base} = \dfrac{XZ}{XY}}}}$

Using phythagoras theorm ${\dag}$

➨ (Hypotenuse)² = (Base)² + (Perpendicular)²

➨ (XZ)² = (XY)² + (YZ)²

➨ (2)² = ( ${\bold{\sf{\sqrt{3}}}}$ )² + (YZ)²

➨ 4 = ${\bold{\sf{\sqrt{3}}}}$ × ${\bold{\sf{\sqrt{3}}}}$ + (YZ)²

➨ 4 = 3 + (YZ)²

➨ 4 - 3 = (YZ)²

➨ 1 = (YZ)²

➨ √1 = YZ

➨ YZ = √1

Now according to the question ${\dag}$

➨ ${\bold{\sf{\dfrac{1-cos \theta}{1+cos \theta}}}}$

➨ 1 - XY / XZ / 1 + XY / XZ

➨ 1 - √3/2/1 + √3/2

➨ 2 - √3/2/2 + √3/2

➨ 2 - √3/2 × 2/2 + √3

➨ 2 - √3/2 + √3

Now let's relationalize ${\dag}$

➨ (2-√3)² / (2)² - (√3)²

➨ (2)² + (√3)² - 2 × 2 × √3/4 - 3

➨ 4 + 3 - 4√3 / 1

➨ 7 - 4√3 / 1

➨ 7 - 4√3

• Henceforth, 7-4√3 is the value of ${\bold{\sf{\dfrac{1-cos \theta}{1+cos \theta}}}}$