Question

prove that tan²A(1+cot²A)/(1+tan²A) = 1​

Answer 1

$\\ \large{ \underline { \underline{ \textbf{Question}}}}$

$\\ \tt{ \ratio{ \rightarrow{Prove \: that - {tan}^{2} A \: \dfrac{(1 + {cot}^{2}A) }{(1 + {tan}^{2}A) } }}}$

$\\ \large{ \underline { \underline{ \textbf{Required \: Answer}}}}$

$\\ \underline{ \underline{ \sf{ \frak{Formula \: Needed \ratio - }}}}$

• $\sf{ tan\theta = \dfrac{sin \theta}{cos \theta}}$

• $\sf{cot \theta = \dfrac{cos \theta}{sin \theta}}$

$\\ \underline{ \underline{ \sf{ \frak{Solution \ratio - }}}}$

$\\ \sf{ \implies{\green{L.H.S.}}}$

$\\ \bold{ {tan}^{2} A \: \dfrac{(1 + {cot}^{2}A) }{(1 + {tan}^{2}A) }}$

$\\ \tt{ \implies{ \dfrac{ {sin}^{2}A }{ {cos}^{2}A } \times \dfrac{1 + \dfrac{ {cos}^{2} A}{ {sin}^{2} A} }{1 + \dfrac{ {sin}^{2} A}{ {cos}^{2}A} }}}$

$\\ \tt{ \implies{ \dfrac{ {sin}^{2}A }{ {cos}^{2}A } \times \dfrac{ \dfrac{ {sin}^{2}A + {cos}^{2}A}{ {cos}^{2}A } }{ \dfrac{ {cos}^{2}A + {sin}^{2}A }{ {sin}^{2}A } }}}$

$\\ \tt{ \implies{ \dfrac{ \cancel{{sin}^{2}A }}{ \cancel{{cos}^{2}A} } \times \dfrac{ \cancel{( {sin}^{2}A + {cos}^{2} A)} }{ \cancel{{sin}^{2}A }} } \times \dfrac { \cancel{{cos}^{2}A } }{ \cancel{({sin}^{2}A+ {cos}^{2}A)}} }$

$\\ \tt{ \implies{ \red{1}}}$

$\\ \sf{ \implies{ \green{R.H.S.}}}$

$\\ \rm{ \therefore \: \tt{ {tan}^{2} A \: \dfrac{(1 + {cot}^{2}A) }{(1 + {tan}^{2}A)} = 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{ \pmb{Proved!!}}}}}$

Answer 2

Step-by-step explanation:

$\begin{gathered} \\ \large{ \underline { \underline{ \textbf{Question}}}}\end{gathered}$

Question

$\begin{gathered} \\ \tt{ \ratio{ \rightarrow{Prove \: that - {tan}^{2} A \: \dfrac{(1 + {cot}^{2}A) }{(1 + {tan}^{2}A) } }}}\end{gathered}$

:→

$\begin{gathered} \\ \large{ \underline { \underline{ \textbf{Required \: Answer}}}}\end{gathered}$

$\begin{gathered} \\ \underline{ \underline{ \sf{ \frak{Formula \: Needed \ratio - }}}}\end{gathered}$

:−

\

$\begin{gathered}\\ \tt{ \implies{ \dfrac{ \cancel{{sin}^{2}A }}{ \cancel{{cos}^{2}A} } \times \dfrac{ \cancel{( {sin}^{2}A + {cos}^{2} A)} }{ \cancel{{sin}^{2}A }} } \times \dfrac { \cancel{{cos}^{2}A } }{ \cancel{({sin}^{2}A+ {cos}^{2}A)}} }\end{gathered}$