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

Question

Q)  Prove that : tan θ /(1 – cot θ) + cot θ / (1 - tan θ) = 1 + tan θ + cot θ

Sapling Academy · Accepted Answer

Here, let's start by simplifying the LHS in given equation:

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

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

= $ \frac{	an ^2 	heta}{	an	heta - 1} + \frac{\frac{1}{	an 	heta}}{1 - 	an 	heta}$

Let's make the denominators equal:

LHS = $ \frac{	an ^2 	heta}{	an	heta - 1} - \frac{\frac{1}{	an 	heta}}{	an 	heta - 1}$

= $\frac{	an ^2 	heta - \frac{1}{	an 	heta}}{	an 	heta - 1}$

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

We know that a3 - b3 = (a - b) (a2 + b2 + a b)

Hence, tan3  θ  - 1 = tan3  θ - 13  = (tan θ - 1) (tan2  θ + 1 + tan θ )

By substituting this value in nominator of LHS, we get:

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

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

= $\frac{	an ^2 	heta}{	an 	heta} + \frac{1}{	an 	heta} + \frac {	an 	heta} {	an 	heta}$

= tan θ + cot θ  + 1

= 1 + tan θ + cot θ = RHS

Hence Proved !

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