Question

If $cot\Theta=\frac{1}{√3}$, write the value of$\frac{1-cos^{2}\Theta}{1-sin^{2}\Theta}$.

Answer 1

SOLUTION :  

Given : cot θ = 1 /√3

cot θ = Base / perpendicular = 1/√3

cot θ = 1/√3

Hypotenuse = √( perpendicular)² + (Base)²

[By Pythagoras theorem]

Hypotenuse = √ (√3² + 1²) = √ 3 + 1 = √4 = 2

Hypotenuse = 2

cos θ  = Base/ hypotenuse = ½  

cos θ  = 1/2

sin θ = perpendicular/hypotenuse = √3/2

sin θ = √3/2

The value of : 1 - cos²θ / 2 - sin²θ

= [ 1 -( ½)²] / [2 - (√3/2)²]

= [ 1 - ¼] / [ 2 - ¾]

= [(4 - 1)/4] / [( 2×4 - 3)/4]

= (¾) / [(8 - 3)/4]

= (¾) / (5/4)

= ¾ × ⅘

1 - cos²θ / 2 - sin²θ = ⅗  

Hence, the value of  1 - cos²θ / 2 - sin²θ is ⅗ .

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Answer 2

Solution:

Identities used

$1 + {cot}^{2}\Theta = {cosec}^{2} \Theta \\ \\ 1 + {tan}^{2} \Theta = {sec}^{2}\Theta$

$\frac{1-cos^{2}\Theta}{1-sin^{2}\Theta}$

Divide numerator and denominator by
${si n}^{2} \Theta$

$\frac{ \frac{1-cos^{2}\Theta}{ {sin}^{2}\Theta}}{ \frac{1-sin^{2}\Theta}{sin^{2}\Theta}} \\ \\ = \frac{ {cosec}^{2}\Theta - {cot}^{2} \Theta }{ {cosec}^{2}\Theta - 1 } \\ \\ = \frac{1 + {cot}^{2}\Theta - {cot}^{2} \Theta }{ 1 + {cot}^{2}\Theta - 1 } \\ \\ = \frac{1}{ {cot}^{2}\Theta } \\ \\ = \frac{1}{ \frac{1}{3} } \\ \\\frac{1-cos^{2}\Theta}{1-sin^{2}\Theta} = 3$