Question

༒ Bʀᴀɪɴʟʏ Sᴛᴀʀs

༒Mᴏᴅᴇʀᴀᴛᴇʀs

༒Other best user

Proof that.

If you don't know then then don't give
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Answer 1

Answer: Proof  is below.

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Step-by-step explanation:

$\sqrt{\frac{1-cos \alpha }{1+ cos \alpha } }$  

= $\sqrt{\frac{(1-cos \alpha )(1+cos\alpha )}{(1+cos\alpha )(1+cos\alpha )} }$              -----> rationalizing numerator

= √ ( 1 -cos²$\alpha$)  / ( 1+ cos $\alpha$)

= √sin² $\alpha$   /  ( 1+ cos $\alpha$)

= sin $\alpha$ / ( 1+ cos $\alpha$)

proved.

Answer 2

Answer:

⠀⠀⠀⌬⠀Prove That :

$\begin{gathered}\qquad \bigstar \:\: \pmb{\sf \sqrt{\dfrac{ 1 - Cos \:\big( \alpha \big) }{ 1 + Cos \big( \alpha \:\big) } }\:\:=\:\: \dfrac{ sin \:\big( \alpha \big) }{ 1 + cos \: \big( \alpha \big)}\:}\:\\\\\end{gathered}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

$\begin{gathered}\qquad \dashrightarrow \sf \sqrt{\dfrac{ 1 - Cos \:\big( \alpha \big) }{ 1 + Cos \big( \alpha \:\big) } }\:\:=\:\: \dfrac{ sin \:\big( \alpha \big) }{ 1 + cos \: \big( \alpha \big)\:} \:\\\\ \end{gathered}$

$\begin{gathered}\qquad \bigstar \:\:\underline {\pmb{\sf{\purple { By \:\:Taking \:\:L.H.S \:\::\:}}}}\\\\\end{gathered}$

$\begin{gathered} \qquad \dashrightarrow \sf \sqrt{\dfrac{ 1 - Cos \:\big( \alpha \big) }{ 1 + Cos \big( \alpha \:\big) } }\:\: \:\\\\\\\end{gathered}$

$\begin{gathered}\qquad \dashrightarrow \sf \sqrt{\dfrac{ 1 - Cos \:\big( \alpha \big) }{ 1 + Cos \big( \alpha \:\big) }} \:\times \: \sqrt{\dfrac{ 1 - Cos \:\big( \alpha \big) }{ 1 + Cos \big( \alpha \:\big) }} \:\: \:\\\\\\\end{gathered}$

$\begin{gathered} \qquad \dashrightarrow \sf \sqrt{\dfrac{\Big\{ 1 - Cos \:\big( \alpha \big)\:\Big\} \Big\{ 1 - Cos \big( \alpha \:\big) \Big\} }{ 1 + Cos \big( \alpha \:\big)\Big\{ 1 - Cos \big( \alpha \:\big) \Big\} } }\:\: \:\\\\\\\end{gathered}$

$\begin{gathered} \qquad \dashrightarrow \sf \sqrt{\dfrac{ 1^2 - Cos^2 \:\big( \alpha \big) }{ \Big\{ 1 + Cos \big( \alpha \:\big) \Big\}^2 }} \:\:\qquad \:\because \:\bigg\lgroup \sf{ ( a^2 - b^2 )\:=\: ( a + b ) \:( a - b) }\bigg\rgroup\: \\\\\\\end{gathered}$

$\begin{gathered}\qquad \dashrightarrow \sf \sqrt{\dfrac{ 1 - Cos^2 \:\big( \alpha \big) }{ \Big\{ 1 + Cos \big( \alpha \:\big) \Big\}^2 } }\:\: \:\\\\\\\end{gathered}$

$\begin{gathered}\qquad \dashrightarrow \sf \sqrt{\dfrac{ Sin^2 \:\big( \alpha \big) }{ \Big\{ 1 + Cos \big( \alpha \:\big) \Big\}^2 } }\:\: \qquad \:\because \:\bigg\lgroup \sf{ \:1\:-\:Cos^2\theta \: \:=\:sin^2\:\theta }\bigg\rgroup \:\\\\\\\end{gathered}$

$\begin{gathered}\qquad \dashrightarrow \sf \dfrac{ Sin \:\big( \alpha \big) }{ 1 + Cos \big( \alpha \:\big) } \:\: \:\\\\\\\end{gathered}$

$\begin{gathered}\qquad \dashrightarrow \:\: \purple {\pmb{\sf \sqrt{\dfrac{ 1 - Cos \:\big( \alpha \big) }{ 1 + Cos \big( \alpha \:\big) } }\:\:=\:\: \dfrac{ sin \:\big( \alpha \big) }{ 1 + cos \: \big( \alpha \big)}\:}}\:\:\bigstar \\\\\end{gathered}$

$\begin{gathered}\qquad \therefore \:\:\underline {\pmb{ \bf \: Hence \:,\:Verified \:\:!\:}}\:\\\\\end{gathered}$

Thanks.