Math, asked by chinmaybhatia9, 3 days ago

Prove the trigonometry identity

(1 + cosx + sinx)^{2}  = 2(1 + sinx)(1 + cosx) \\

Answers

Answered by mathdude500
3

Step-by-step explanation:

Consider,

\sf \: {(1 + cosx + sinx)}^{2}  \\  \\

\sf \: = {(1)}^{2} +  {(cosx)}^{2} +  {(sinx)}^{2} + 2(1)(cosx) + 2(cosx)(sinx) + 2(sinx)(1) \\  \\

\boxed{ \sf{ \: \because \:  {(x + y + z)}^{2} =  {x}^{2} +  {y}^{2} +  {z}^{2} + 2xy + 2yz + 2zx \: }}

\sf \: =  \: 1 +  {cos}^{2}x +  {sin}^{2}x + 2cosx + 2sinxcosx + 2sinx \\  \\

\sf \: =  \: 1 + 1 + 2cosx + 2sinxcosx + 2sinx \\  \\

\boxed{ \sf{ \: \because \:  {sin}^{2}x +  {cos}^{2}x = 1 \: }} \\  \\

\sf \: =  \: 2 + 2cosx + 2sinxcosx + 2sinx \\  \\

\sf \: =  \: (2 + 2cosx) + (2sinxcosx + 2sinx) \\  \\

\sf \: =  \: 2(1 + cosx) + 2sinx(cosx + 1) \\  \\

\sf \: =  \: 2(1 + cosx) + 2sinx(1 + cosx) \\  \\

\sf \: =  \: 2(1 + cosx)(1 + sinx) \\  \\

Hence,

\boxed{ \sf{ \:\bf \: {(1 + cosx + sinx)}^{2} = 2(1 + sinx)(1 + cosx) \: }}\\  \\

\rule{190pt}{2pt}

Additional Information

\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sinx =  \dfrac{1}{cosecx} }\\ \\ \bigstar \: \bf{cosx =  \dfrac{1}{secx} }\\ \\ \bigstar \: \bf{tanx = \dfrac{sinx}{cosx}  = \dfrac{1}{cotx} }\\ \\ \bigstar \: \bf{cot x= \dfrac{cosx}{sinx}  = \dfrac{1}{tanx} }\\ \\ \bigstar \: \bf{cosec x) = \dfrac{1}{sinx} }\\ \\ \bigstar \: \bf{secx = \dfrac{1}{cosx} }\\ \\ \bigstar \: \bf{ {sin}^{2}x +  {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x -  {tan}^{2}x = 1  }\\ \\ \bigstar \: \bf{ {cosec}^{2}x -  {cot}^{2}x = 1 } \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}

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