Question

$\binom{ \tan( \alpha ) }{1 - \cot( \alpha =) } + \binom{ \cot( \alpha ) }{1 - \tan( \alpha ) } = 1 + \sec( \alpha) \: \csc( \alpha ) )$
Prove this Identity, where the angles involved are acute angles
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Answer 1

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$\begin{lgathered}\implies\textbf{\red{T}\blue{o}\:\green{p}\purple{r}\orange{o}\pink{v}e}:\\\boxed{\bf\dfrac{\tan(A)}{1-\cot(A)}+\dfrac{\cot(A)}{1-\tan(A)}=1+\sec(A)\csc(A)}\end{lgathered}$

$\begin{lgathered}\bigstar\underline{\red{\bf\:Proof:}}\bigstar\\\red\leadsto \bf\dfrac{\tan(A) }{1-\cot(A)}+\dfrac{\cot(A)}{1-\tan(A)}\end{lgathered}$

Putting tanA = (sinA/cosA) & cotA = (cosA/sinA) we get,

$\begin{lgathered}\red\leadsto\dfrac{\frac{\sin(A)}{\cos(A)}}{1-\frac{\cos(A)}{ \sin(A)}} + \dfrac{\frac{ \cos(A) }{\sin(A)}}{1 - \frac{\sin(A)}{\cos(A)}} \\ \\ \red\leadsto\dfrac{\frac{\sin(A)}{\cos(A)}}{\frac{\sin(A)-\cos(A)}{\sin(A)}}+\dfrac{ \frac{\cos(A)}{\sin(A)}}{\frac{\cos(A)-\sin(A)}{\cos(A)}} \\ \\\red\leadsto\:\dfrac{\sin^{2} (A) }{\cos(A)(\sin(A) - \cos(A))}+\dfrac{\cos^{2}(A)}{\sin(A)(\cos(A)-\sin(A))}\\\\\green{\textbf{Taking Negative Common from denominator}}\\\\ \red\leadsto\dfrac{\sin^{2}(A)}{\cos(A)(\sin(A)-\cos(A))}-\dfrac{\cos^{2}(A)}{\sin(A)(\sin(A)-\cos(A))} \\ \\ \red\leadsto \dfrac{1}{( \sin(A) - \cos(A))}\bigg(\dfrac{\sin^{2}(A)}{\cos(A)}-\dfrac{\cos^{2}(A)}{\sin(A)} \bigg) \\ \\\red\leadsto\dfrac{1}{(\sin(A) - \cos(A))}\bigg(\dfrac{\sin^{3}(A)-\cos ^{3}(A)}{\cos(A)\sin(A)}\bigg)\end{lgathered}$

$\green{\bf\:using}:\pink{\boxed{\bf \: {a}^{3}-{b}^{3}=(a-b)({a}^{2}+{b}^{2}+ab)}}$

$\begin{lgathered}\red\leadsto\dfrac{1}{\cancel{(\sin(A) - \cos(A))}}\times \dfrac{\cancel{(\sin(A) - \cos(A))}(\sin^{2}(A)+\cos^{2}(A)+ \sin(A)\cos(A))}{\cos(A)\sin(A)} \\ \\ \red\leadsto\dfrac{(\sin^{2}(A)+\cos ^{2}(A)+\sin(A)\cos(A))}{\cos(A)\sin(A)} \\\\\purple{\boxed{\bf\:\sin^{2}(A)+\cos^{2}(A)=1}}\\\\ \red\leadsto\dfrac{(1 +\sin(A)\cos(A))}{\cos(A)\sin(A)}\\\\\red\leadsto \dfrac{1}{\cos(A)\sin(A)}+ \cancel\dfrac{\sin(A)\cos(A) }{\cos(A)\sin(A)}\\\\\red\leadsto\dfrac{1}{\cos(A)\sin(A)}+1\end{lgathered}$

$\begin{lgathered}\blue{\boxed{\bf\dfrac{1}{\cos(A)} =\sec(A)}}\\ \orange{\boxed{\bf\dfrac{1}{\sin(A)}=\cosec(A)}}\end{lgathered}$

$\red\leadsto\:\:\red\therefore\: \red{\large\boxed{\bf1+\sec(A)\csc(A)}}$

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Answer 2

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$\huge\tt{PROBLEM:}$

$\binom{ \tan( \alpha ) }{1 - \cot( \alpha =) } + \binom{ \cot( \alpha ) }{1 - \tan( \alpha ) } = 1 +y \sec( \alpha) \: \csc( \alpha ) )$

Prove this Identity, where the angles involved are acute angles

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$\huge\tt{SOLUTION:}$

=tanA/(1–1/tanA) +cotA/(1-tanA)

=tan^2A/(tanA-1) +cotA/(1-tanA)

=-tan^2A/(1-tanA) +cotA/(1-tanA)

=(-tan^2A+cotA)/(1-tanA)

=(-tan^2A+1/tanA)/(1-tanA)

=(-tan^3A+1)/tanA(1-tanA)

=(1-tan^3A)/tanA(1-tanA)

=(1-tanA)(1+tanA+tan^2A)/tanA(1-tanA)

=(1+tanA+tan^2A)/tanA

=(1+tan^2A+tanA)/tanA

=(sec^2A+tanA)/tanA

=sec^2A/tanA +tanA/tanA

=1/cos^2AtanA+1

=cosA/cos^2AsinA+1

=1/cosAsinA+1

=secA cosecA + 1

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