Question

Prove the following question:

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

LHS = RHS

Step-by-step explanation:

Given that,

$\sf1 + \ \dfrac{cot^{2}\theta}{1 \ + \ cosec\theta} = \dfrac{1}{sin\theta}$

$\sf LHS = 1 \ + \ \dfrac{cot^{2}\theta}{1 \ + \ cosec\theta}$

$\boxed{\sf cot^{2}\theta = cosec^{2}\theta - 1}$

$\sf LHS = 1 \ + \ \dfrac{cosec^{2}\theta - 1}{1 \ + \ cosec\theta}$

$\sf LHS = 1 \ + \ \dfrac{(cosec\theta - 1)(cosec\theta + 1)}{1 \ + \ cosec\theta}$

$\sf (cosec\theta + 1 \ and \ cosec\theta + 1 \ in \ the \ numerator \ and \ the \ denominator \ get \ canceled.)$

$\sf LHS = 1 \ + \ cosec\theta - 1$

$\sf LHS = cosec\theta$

$\large\boxed{\sf{cosec\theta = \frac{1}{sin\theta}}}$

$\sf\Longrightarrow LHS = \dfrac{1}{sin\theta}$

Hence Proved!