Question

Sin theta divided by 1 - cos theta is equal to cosec theta + cot theta.​

Answer 1

Question:

sin∅/ (1-cos∅) = cosec∅ + cot∅

Solution:

∵       LHS = $\frac{sin∅}{(1-cos∅)}$

=>     LHS = $\frac{sin∅(1+cos∅)}{(1-cos∅) (1+cos∅)}$

=>     LHS = $\frac{sin∅+cos∅sin∅}{(1-cos∅^{2} )}$

=>     LHS = $\frac{sin∅}{1-cos∅^{2} } +\frac{cos∅sin∅}{1-cos∅^{2}}$

=>     LHS = sin∅/sin²∅ + cos∅sin∅/sin²∅          [∵1-cos²∅= sin²∅]

=>     LHS = 1/sin∅ + cos∅/sin∅

=>     LHS = cosec∅ + cot∅                  [∵1/sin∅=cosec∅ and cos∅/sic∅=tan∅]

∴     LHS = RHS

Hence, Proved

                                                                                   

Hope this helped you..

Answer 2

Answer:

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

Step-by-step explanation:

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

Consider the left hand side of the equation,

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

Dividing and multiplying with ${1+cos\theta}$

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

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

Using the trigonometric identity, $sin^{2} \theta}+cos^{2} \theta=1$

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

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

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

Using the relations,  $\frac{1 }{sin \theta}= cosec \theta ~~\&~~\frac{cos\theta }{sin \theta}=cot \theta$

$=cosec \theta}+cot\theta$

Therefore, $\frac{sin\theta }{1-cos\theta} =cosec\theta+cot\theta$