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Answer 1

Answer:

Step-by-step explanation:

Given:

$\sf \dfrac{sin\: \theta}{cot\: \theta + cosec\: \theta}=2+\dfrac{sin\: \theta}{cot\: \theta-cosec\: \theta}$

To Prove:

LHS = RHS

Proof:

Consider the LHS of the equation,

$\sf \dfrac{sin\: \theta}{cosec\: \theta + cot\: \theta}$

Multiply numerator and denominator by cosec θ - cot θ

$\sf \implies \dfrac{sin\: \theta(cosec\: \theta-cot\: \theta)}{cosec\: \theta + cot\: \theta(cosec\: \theta-cot\: \theta)}$

We know,

(a + b) (a - b) = a² - b²

$\sf \implies \dfrac{sin\: \theta \times cosec\:\theta-sin\: \theta \times cot\: \theta}{cosec^{2}\: \theta-cot^{2}\: \theta }$

Now we know,

cosec² θ - cot² θ = 1

cosec θ = 1/ sin θ

cot θ = cos θ/sin θ

Applying the identities,

$\sf \implies sin\: \theta \times \dfrac{1}{sin\: \theta} -sin\: \theta \times \dfrac{cos\: \theta}{sin\: \theta} \times \dfrac{1}{1}$

Simplifying,

$\sf \implies 1-cos\: \theta---(1)$

Now considering the RHS of the equation,

$\sf 2+\dfrac{sin\: \theta}{cot\: \theta-cosec\: \theta}$

Multiply both numerator and denominator by cot θ + cosec θ

$\sf \implies 2+\dfrac{sin\: \theta(cot\: \theta+cosec\: \theta)}{cot\: \theta-cosec\: \theta(cot\: \theta+cosec\: \theta)}$

Simplifying,

$\sf \implies 2+\dfrac{sin\: \theta cot\: \theta+sin\: \theta cosec\: \theta}{cot^{2}\: \theta-cosec^{2}\: \theta }$

Here,

cot² θ - cosec² θ = -1

Apply the identities,

$\sf \implies 2 + \dfrac{cos\: \theta+ 1}{-1}$

$\sf \implies 2 + -(cos\: \theta+ 1)$

$\sf \implies 2 + -cos\: \theta- 1$

$\sf \implies 1 -cos\: \theta----(2)$

From equations 1 and 2, RHS are equal, hence LHS must also be equal.

That is,

$\sf \dfrac{sin\: \theta}{cot\: \theta + cosec\: \theta}=2+\dfrac{sin\: \theta}{cot\: \theta-cosec\: \theta}$

Hence proved.