Math, asked by durgaprasaddasyam, 10 months ago

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Answered by gitanjali1013
4

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Answered by RvChaudharY50
76

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

\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)}

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

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

\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)

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

 \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

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

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

\rule{200}{4}

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


Ridvisha: wow....xd
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