Question

The value of Cot theta+ Cosec theta -1/cot theta-cosec theta+1​

Answer 1

Given :

$\bf\dfrac{\cot\theta+\cosec\theta-1}{\cot\theta-cosec\theta+1}$

Explanation :

We know that,

• 1 = cosec²θ - cot²θ

Replacing 1,

$\\ =\sf\dfrac{\cot\theta+\cosec\theta-(\cosec^2\theta-\cot^2\theta)}{\cot\theta-\cosec\theta+1}$

• cosec²θ - cot²θ = (cosecθ + cotθ)(cosecθ - cotθ)

$\\ =\sf\dfrac{\cot\theta+\cosec\theta-[(\cosec\theta+\cot\theta)(\cosec\theta-\cot\theta)]}{\cot\theta-\cosec\theta+1}$

Taking cotθ + cosecθ as common,

$\\ =\sf\dfrac{\cot\theta+\cosec\theta[1-(\cosec\theta-\cot\theta)]}{\cot\theta-\cosec\theta+1}$

$\\ =\sf\dfrac{\cot\theta+\cosec\theta[1-\cosec\theta+\cot\theta]}{\cot\theta-\cosec\theta+1}$

Cancelling cotθ - cosecθ + 1,

$\\ =\sf\dfrac{\cot\theta+\cosec\theta\ [\cancel{1-\cosec\theta+\cot\theta}]}{\cancel{\cot\theta-\cosec\theta+1}}$

$\\ =\sf\cot\theta+\cosec\theta$

• cotθ = cosθ/sinθ
• cosecθ = 1/sinθ

Replacing cotθ & cosecθ,

$\\ =\sf\dfrac{\cos\theta}{\sin\theta}+\dfrac{1}{\sin\theta}$

$\\ =\sf\dfrac{\cos\theta+1}{\sin\theta}$

$\therefore\boxed{\bf\dfrac{\cot\theta+\cosec\theta-1}{\cot\theta-cosec\theta+1}=\dfrac{\cos\theta+1}{\sin\theta}}$