Prove the; (tan 𝛂 + ((cos 𝛂)/(1 + sin 𝛂))) sin 𝛂 = 𝛂 ^([Z.A])


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Question Number 164162 by Zaynal last updated on 15/Jan/22

$Prove the;$ $(t a n α + \frac{c o s α}{1 + s i n α}) s i n α = α$ $^{[Z . A]}$
Commented by som(math1967) last updated on 15/Jan/22
${tan𝛂 +((cos𝛂(1−sin𝛂))/(1−sin^2 𝛂))}×sinα ={tan𝛂+((cos𝛂(1−sin𝛂))/(cos^2 𝛂))}×sin𝛂 ={tan𝛂+(1/(cos𝛂)) −tan𝛂}×sin𝛂 =sec𝛂×sin𝛂=tan𝛂 so (tan𝛂+((cos𝛂)/(1+sin𝛂)))sin𝛂=tan𝛂 ≠𝛂$
${t a n α + \frac{c o s α (1 - s i n α)}{1 - {s i n}^{2} α}} \times s i n α$ $= {t a n α + \frac{c o s α (1 - s i n α)}{{c o s}^{2} α}} \times s i n α$ $= {t a n α + \frac{1}{c o s α} - t a n α} \times s i n α$ $= s e c α \times s i n α = t a n α$ $s o (t a n α + \frac{c o s α}{1 + s i n α}) s i n α = t a n α$ $\neq α$
Commented by Zaynal last updated on 15/Jan/22

$thank you sir . .,$ $answer is \tan apha$
Commented by cortano1 last updated on 15/Jan/22

Commented by som(math1967) last updated on 15/Jan/22

$s o r r y t y p o i c o r r e c t e d$
Commented by Zaynal last updated on 15/Jan/22

$thank you sir, very good .$
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