Can anyone solve this plz

1.	Can anyone solve this plz
Answer» To Find: $\Longrightarrow \sf \dfrac{sinA}{1 + cosA} + \dfrac{1 + cosA}{sinA}$ Taking LCM we get: $\Longrightarrow \sf \dfrac{(sinA)(sinA) + (1 + cosA)(1 + cosA)}{(1 + cosA)(sinA)}$ Using (a + B)(a + b) = (a + b)² we get: $\Longrightarrow \sf \dfrac{sin^2A + (1 + cosA)^2}{(1 + cosA)(sinA)}$ Using (a + b)² = a² + b² + 2ab we get: $\Longrightarrow \sf \dfrac{sin^2A + (1)^{2} + (cosA)^2 + 2(1)(cosA)}{(1 + cosA)(sinA)}$ $\Longrightarrow \sf \dfrac{sin^2A + 1 + cos^2A + 2cosA}{(1 + cosA)(sinA)}$ $\Longrightarrow \sf \dfrac{sin^2A + cos^2A + 1 + 2cosA}{(1 + cosA)(sinA)}$ Using the identity sin²A + cos²A = 1 we get: $\Longrightarrow \sf \dfrac{1 + 1 + 2cosA}{(1 + cosA)(sinA)}$ $\Longrightarrow \sf \dfrac{2 + 2cosA}{(1 + cosA)(sinA)}$ TAKE 2 out since it's common to both 2 and 2cosA. $\Longrightarrow \sf \dfrac{2(1 + cosA)}{(1 + cosA)(sinA)}$ Cancelling (1 + cosA) in the numerator & DENOMINATOR we get: $\Longrightarrow \sf \dfrac{2}{sinA}$ $\Longrightarrow \sf 2 \times \dfrac{1}{sinA}$ Using 1/sinA = cosecA $\Longrightarrow \sf 2cosecA$ Hence solved.

Answer»

To Find:

$\Longrightarrow \sf \dfrac{sinA}{1 + cosA} + \dfrac{1 + cosA}{sinA}$