Prove that: sin θ/(cot θ + cosec θ) = 2 + sin θ /(cot θ

Q). Prove that: sin θ/(cot θ + cosec θ) = 2 + sin θ /(cot θ – cosec θ)

Ans. Let’s start from LHS:

Step 1: LHS = $\frac{\sin \theta}{\cot \theta + \cos ec \theta}$

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

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

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

Step 2: We know that sin² θ + cos² θ = 1

∴ sin² θ = 1 – cos² θ

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

Step 3: We know that a² – b² = (a + b) ( a – b)

1 – cos² θ = 1² – cos² θ = (1 + cos θ ) ( 1- cos θ )

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

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

= 1 – cos θ ………… (i)

Step 4: Next we take RHS:

RHS = 2 + $\frac{\sin \theta}{\cot \theta - \cos ec \theta}$

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

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

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

Step 5: Since, sin² θ = 1 – cos² θ

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

Step 6: Since 1 – cos² θ = (1 + cos θ ) ( 1- cos θ )

∴ RHS = 2 + $\frac{(1 - \cos \theta)(1 + \cos \theta)}{(\cos \theta - 1)}$

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

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

= 2 – (1 + cos θ)

= 2 – 1 – cos θ

= 1 – cos θ .……….. (ii)

Step 7: By comparing (i) and (ii), we get: LHS = RHS ………. Hence Proved !

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