Question

$1-\frac{Si{n}^{2}y}{1+Cosy}+\frac{1+Cosy}{Siny}-\frac{Siny}{1-Cosy}=........$

• A. $0$
• B. $1$
• C. $\sin y$
• D. $\cos y$

Answer 1

The correct option is D $\cos y$

$1-\frac{si{n}^{2}y}{1+cosy}+\frac{1+cosy}{siny}-\frac{siny}{1-cosy}$
$\equiv \frac{1+cosy-si{n}^{2}y}{1+cosy}+\frac{1-co{s}^{2}y-si{n}^{2}y}{siny(1-cosy)}$
$\equiv \frac{cosy+co{s}^{2}y}{1+cosy}+\frac{1+(si{n}^{2}y+co{s}^{2}y)}{sin\left(y\right)(1-cosy)}$
$\equiv \frac{cosy(1+cosy)}{1+cosy}+\frac{1-1}{siny(1-cosy)}$
$\equiv cosy$