Question

If A =30 degree prove 1 + sin 2 A + cos 2 A/
sin A +cos A=2cosA​

Answer 1

Given Question:

If A = 30° prove that $\sf\dfrac{1+ sin \: 2A + cos\: 2A}{sin\:A + cos\:A} = 2 \: cos\: A$

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ᴛ ᴀ ᴋ ɪ ɴ ɢ ʟ ʜ s :

$:\implies\sf\dfrac{1+ sin \: 2A + cos\: 2A}{sin\:A + cos\:A} \\\\\\:\implies\sf \dfrac{1 + sin \: 2 \Big(30^{\circ}\Big) + cos \: 2 \Big(30^{\circ} \Big)}{sin\: 30^{\circ} + cos \: 30^{\circ}} \\\\\\:\implies\sf \dfrac{1 + \dfrac{\sqrt{3}}{\:2} + \dfrac{1}{2}}{\dfrac{1}{2}+ \dfrac{\sqrt{3}}{\;2}} \\\\\\:\implies\sf \dfrac{3 + \sqrt{3}}{\sqrt{3} + 1} \Bigg[ \dfrac{\sqrt{3} - 1}{\sqrt{3} - 1} \Bigg]\\\\\\:\implies\sf \dfrac{3 \:\sqrt{3} - 3 + 3 - \sqrt{3}}{2}\\\\\\:\implies{\underline{\boxed{\frak{\pink{\sqrt{3}}}}}}\:\bigstar$

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Now,

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ᴛ ᴀ ᴋ ɪ ɴ ɢ ʀ ʜ s :

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$:\implies\sf 2 \: cos \: A \\\\\\:\implies\sf 2 \: cos \: \Big(30^{\circ}\Big) \\\\\\:\implies\sf 2 \Bigg[ \dfrac{\sqrt{3}}{2} \Bigg]\\\\\\:\implies{\underline{\boxed{\frak{\pink{\sqrt{3}}}}}}\:\bigstar$

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ʜᴇɴᴄᴇ ᴘʀᴏᴠᴇᴅ!

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$\bigstar\:\sf Trigonometric\:Values :\\\begin{tabular}{|c|c|c|c|c|c|}\cline{1-6}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}& 0\\\cline{1-6}Tan \theta& 0& \dfrac{1}{\sqrt{3}}&1& \sqrt{3}& Not D \hat{e} fined \\\cline{1-6}\end{tabular}$

Answer 2

Given :-

• A = 30°

To Prove :-

$\bf\dfrac{1+ sin \: 2A + cos\: 2A}{sin\:A + cos\:A} = 2 \: cos\: A$

Solution :-

★ Taking L.H.S ★

$\begin{gathered}\longmapsto\sf\dfrac{1+ sin \: 2A + cos\: 2A}{sin\:A + cos\:A} \\\\\\\longmapsto\sf \dfrac{1 + sin \: 2 \Big(30^{\circ}\Big) + cos \: 2 \Big(30^{\circ} \Big)}{sin\: 30^{\circ} + cos \: 30^{\circ}} \\\\\\\longmapsto\sf \dfrac{1 + \dfrac{\sqrt{3}}{\:2} + \dfrac{1}{2}}{\dfrac{1}{2}+ \dfrac{\sqrt{3}}{\;2}} \\\\\\\longmapsto\sf \dfrac{3 + \sqrt{3}}{\sqrt{3} + 1} \Bigg( \dfrac{\sqrt{3} - 1}{\sqrt{3} - 1} \Bigg)\\\\\\\longmapsto\sf \dfrac{3 \:\sqrt{3} - 3 + 3 - \sqrt{3}}{2}\\\\\\\longmapsto{\underline{\sqrt{3}}}\:\end{gathered}$

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★ R.H.S = 2 cos A ★

⠀$\begin{gathered}{:} \longrightarrow\sf 2 \: cos \: A \\\\\\{:} \longrightarrow\sf 2 \: \Big(30^{\circ}\Big) \\\\\\{:} \longrightarrow\sf 2 \Bigg( \dfrac{\sqrt{3}}{2} \Bigg) \\ \\ \\ \longrightarrow\sf \cancel2 \: \times \dfrac{\sqrt{3}}{ \cancel2} \\\\\\{:} \longrightarrow{\underline{\sqrt{3}}}\:\end{gathered}$

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RHS = LHS ( √3 )

★ Hence, Proved★

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★ Additional Info :-

Trigonometric Full Table :

$\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}$

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