Prove-the-tan-cos-1-sin-sin-Z-A-

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

{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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