frac{tan theta}{1-cot theta}+frac{cot theta}{ | vedclass

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Question

(D) The given expression is $\frac{	an 	heta}{1-\cot 	heta}+\frac{\cot 	heta}{1-	an 	heta}$.Substitute $\cot 	heta = \frac{1}{	an 	heta}$:$=

Vedclass · Accepted Answer

$1+	an 	heta+\cot 	heta$

Vedclass · Answer

(D) The given expression is $\frac{	an 	heta}{1-\cot 	heta}+\frac{\cot 	heta}{1-	an 	heta}$.Substitute $\cot 	heta = \frac{1}{	an 	heta}$:$= \frac{	an 	heta}{1-\frac{1}{	an 	heta}} + \frac{\frac{1}{	an 	heta}}{1-	an 	heta} = \frac{	an^2 	heta}{	an 	heta - 1} + \frac{1}{	an 	heta(1-	an 	heta)}$.Rewrite the second term to have a common denominator $(	an 	heta - 1)$:$= \frac{	an^2 	heta}{	an 	heta - 1} - \frac{1}{	an 	heta(	an 	heta - 1)}$.Combine the fractions:$= \frac{	an^3 	heta - 1}{	an 	heta(	an 	heta - 1)}$.Use the algebraic identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ where $a = 	an 	heta$ and $b = 1$:$= \frac{(	an 	heta - 1)(	an^2 	heta + 	an 	heta + 1)}{	an 	heta(	an 	heta - 1)}$.Cancel the common term $(	an 	heta - 1)$:$= \frac{	an^2 	heta + 	an 	heta + 1}{	an 	heta} = \frac{	an^2 	heta}{	an 	heta} + \frac{	an 	heta}{	an 	heta} + \frac{1}{	an 	heta}$.$= 	an 	heta + 1 + \cot 	heta = 1 + 	an 	heta + \cot 	heta$.

$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}$ is equal to

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