Prove that: tan θ /(1 – cot θ) + cot θ / (1 – tan θ) = 1 + sec θ cosecθ

Question

Q) Prove that:  $\frac{	an 	heta}{1 - \cot 	heta} + \frac{\cot 	heta}{1 - 	an 	heta}$

Sapling Academy · Accepted Answer

Let’s start from LHS

$\frac{	an 	heta}{1 - \cot 	heta} + \frac{\cot 	heta}{1 - 	an 	heta}$

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

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

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

= $(\frac{1}{\sin 	heta - \cos 	heta}) (\frac{\sin^3 	heta - \cos^3 	heta}{\sin 	heta \cos 	heta})$

We know that, a3−b3 formula is = (a−b)(a2 + b2 + ab)

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

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

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

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

= sec θ  cosec θ + 1

= 1 + sec θ  cosec θ = RHS

Hence Proved

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