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

Question

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

Sapling Academy · Accepted Answer

Let's start from LHS:

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

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

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

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

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

∴ sin2 θ = 1 - cos2 θ

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

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

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

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

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

= 1 - cos θ ............ (i)

Step 4: Next we take RHS:

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

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

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

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

Step 5: Since, sin2 θ = 1 - cos2 θ

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

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

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

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

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

= 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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