Question

proof
I will mark it brainliest​

Answer 1

$\huge{\mathfrak{\overbrace{\underbrace{\pink{\fbox{\green{\blue{\bf\:S}\pink{o}\red{l}\orange{u}\purple{t}\blue{i}\green{o}\red{n}}}}}}}}$

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

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

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

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

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

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

Now,

⟿ (sin²A/sinA*cosA) + (cos²A/sinA*cosA) +(sinA*cosA/sinA*cosA)

⟿ (sinA/cosA) + (cosA/sinA) + 1

$\red\leadsto\boxed{\bf\:\red1\blue+\pink{\tan(A)}+\green{\cot(A)}}\:\:\underline{\boxed{\textbf{\red{P}\blue{r}\green{o}\purple{v}\orange{e}\pink{d}}}}$

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$\bold{\boxed{\huge{\boxed{\orange{\small{\boxed{\huge{\red{\bold{\huge\bold{\red{\ddot{\smile}}}}}}}}}}}}}$