Question

root1-costheta/1+costheta=cosectheta-cottheta​

Answer 1

Answer:

$\sqrt{\frac{1-cos\theta}{1+cos\theta}}=cosec\theta-cot\theta$

Step-by-step explanation:

Formula used:

$a^2-b^2=(a+b)(a-b)$

Now,

$\frac{1-cos\theta}{1+cos\theta}$

$\frac{1-cos\theta}{1+cos\theta}=\frac{1-cos\theta}{1+cos\theta}*\frac{1-cos\theta}{1-cos\theta}$

$\frac{1-cos\theta}{1+cos\theta}=\frac{(1-cos\theta)^2}{1-cos^2\theta}$

$\frac{1-cos\theta}{1+cos\theta}=\frac{(1-cos\theta)^2}{sin^2\theta}$

$\frac{1-cos\theta}{1+cos\theta}=(\frac{1-cos\theta}{sin\theta})^2$

Taking square root on both sides

$\sqrt{\frac{1-cos\theta}{1+cos\theta}}=\frac{1}{sin\theta}-\frac{cos\theta}{sin\theta}$

$\sqrt{\frac{1-cos\theta}{1+cos\theta}}=cosec\theta-cot\theta$

Answer 2

Answer:

Proved

Step-by-step explanation:

√(1-cosθ) / √(1 + cosθ)  = Cosecθ - Cotθ

LHS = √(1-cosθ) / √(1 + cosθ)  

Lets multiply & divide by √(1-cosθ)

= (√(1-cosθ) )² /√(1 - Cos²θ)

= (1- Cosθ)/√Sin²θ

= (1 - Cosθ)/Sinθ

= 1/Sinθ - Cosθ/Sinθ

= Cosecθ - Cotθ

= RHS

QED