Question

please solve this question

Answer 1

$\sqrt{\frac{CosecA + 2CosA}{CosecA - 2CosA}}$

$\sqrt{\frac{\frac{1}{SinA} + 2CosA}{\frac{1}{SinA} - 2CosA}}$

Taking LCM and Cancelling SinA We get :

$\sqrt{\frac{1 + 2SinACosA}{1 - 2SinACosA}}$

We know that Sin²A + Cos²A = 1

Substituting Sin²A + Cos²A in the Place of 1

$\sqrt{\frac{Sin^2A + Cos^2A+ 2SinACosA}{Sin^2A + Cos^2A - 2SinACosA}}$

We know that (a + b)² = a² + b² + 2ab and (a - b)² = a² + b² - 2ab

$\sqrt{(\frac{SinA + CosA}{SinA - CosA})^2}$

$\frac{SinA + CosA}{SinA - CosA}$