Question

Prove :
⠀
$\sf Q. \: \dfrac{sin2 \alpha}{1 + cos \alpha} \times \dfrac{cos \alpha}{1 + cos \alpha} = tan \dfrac{ \alpha}{2}$


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Answer 1

Tσ ρяσνє —

$\sf { \longrightarrow\: \dfrac{sin2 \alpha}{1 + cos2 \alpha} \times \dfrac{cos \alpha}{1 + cos \alpha} = tan \dfrac{ \alpha}{2}}$

ㅤ

Sσℓυтiσи –

$\quad\sf{\longmapsto \left(\dfrac{sin2\alpha }{1+cos2\alpha }\right)\times \left(\dfrac{cos\alpha}{1+cos\alpha}\right)}$

✪ Using Formulas :

$\qquad\quad\sf{\leadsto cos2x = 2 cos^2x-1}$

$\qquad\quad\sf{\leadsto sin2x=2sin\:x\:cos\:x}$

$\qquad\quad\sf{\leadsto sin\:x = 2\:sin\left(\dfrac{x}{2}\right)cos\left(\dfrac{x}{2}\right) }$

$\qquad\quad\sf{\leadsto cos\:x=2cos^2 \left(\dfrac{x}{2}\right) -1}$

ㅤ

Tнєrєfσrє —

$\quad\sf{\longmapsto \left(\dfrac{sin2\alpha }{1+cos2\alpha }\right)\times \left(\dfrac{cos\alpha}{1+cos\alpha}\right)}$

$\quad\sf{\longmapsto \left(\dfrac{2sin\alpha cos\alpha}{2cos^2 \alpha}\right)\left(\dfrac{cos\alpha}{1+cos\alpha}\right)}$

$\quad\sf{\longmapsto \dfrac{2\times sin\alpha\times cos\alpha\times cos\alpha}{2\times cos\alpha\times cos\alpha\times (1+cos\alpha)}}$

$\quad\sf{\longmapsto \dfrac{sin\alpha}{1+cos\alpha}}$

$\quad\sf{\longmapsto \dfrac{2sin\dfrac{\alpha}{2}cos\dfrac{\alpha}{2}}{2cos^2\dfrac{\alpha}{2}}}$

$\quad\sf{\longmapsto \dfrac{\cancel{2}sin\dfrac{\alpha}{2}cos\dfrac{\alpha}{\cancel{2}}}{\cancel{2}\cos\dfrac{\alpha}{2}cos\dfrac{\alpha}{\cancel{2}}}}$

$\quad\sf{\longmapsto \dfrac{sin \left(\dfrac{\alpha}{2}\right)\cancel{cos\alpha}}{cos\left(\dfrac{\alpha}{2}\right)\cancel{cos\alpha}}}$

$\quad\sf{\longmapsto \dfrac{\left(sin\dfrac{\alpha}{2}\right)}{\left(cos\dfrac{\alpha}{2}\right)} }$

$\quad\sf{\longmapsto tan\dfrac{\alpha}{2}}$

❝ Hence , proved ❞