Question

cot theta +cosec theta-1/cot theta-cosec theta +1 = 1+ cos theta/sin theta​

Answer 1

Answer:

Step-by-step explanation:

To Prove :-

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

How To Prove :-

We need substitute a trigonometric identity in place of '1' in the numerator and we need to expand that identity after that we can see that we can take a term as common and we need to simplify it and we need to cancel the common terms. Next we need to change the terms we got in the form of 'sinθ and cosθ'.

Formula Required :-

cosec²θ - cot²θ = 1

cotθ = cosθ/sinθ

cosecθ = 1/sinθ

Solution :-

Taking L.H.S :-

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

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

[ ∴ cosec²θ - cot²θ = 1]

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

[ ∴∴ (a + b)(a - b) = a² - b² ]

Taking 'cosecθ + cotθ' as common :-

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

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

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

Cancelling the common terms :-

= cosecθ + cotθ

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

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

= R.H.S

Hence Proved.