Question

grade 10th it's urgent

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Step-by-step explanation:

$Given:\frac{cosecA-sinA}{cosecA+sinA}$

$=\frac{\frac{1}{sinA}-sinA}{\frac{1}{sinA}+sinA}$

$=\frac{\frac{1-sin^2A}{sinA}}{\frac{1+sin^2A}{sinA}}$

$=\frac{1-sin^2A}{1+sin^2A}$

Divide with cos²A, we get

$=\frac{\frac{1-sin^2A}{cos^2A}}{\frac{1+sin^2A}{cos^2A} }$

$=\frac{\frac{1}{cos^2A}-\frac{sin^2A}{cos^2A}}{\frac{1}{cos^2A}+\frac{sin^2A}{cos^2A}}$

$=\boxed{\frac{sec^2A-tan^2A}{sec^2A+tan^2A}}$

Hope it helps!