Question

prove that (tan + cos + sec ) ( tan + cot - sec ) = cosec^2​

Answer 1

Proof

$(tan (x) + cot(x) + sec(x) ) ( tan(x) + cot(x) - sec(x) )$

$= (tan (x) + cot(x))^2 - sec^{2}(x)$

$= \frac{( sin^2(x) + cos^2(x) )^2}{sin (x) cos (x)} - \frac{1}{cos^2{x}}$

$= \frac{1}{sin^{2} (x) cos^{2} (x)} - \frac{1}{cos^2{x}}$

$= \frac{1 - sin^{2} (x)}{sin^{2} (x) cos^{2} (x)}$

$= \frac{cos^{2} (x)}{sin^{2} (x) cos^{2} (x)}$

$= cosec^{2}(x)$

Used formulas

$a^{2}-b^{2} = (a - b) (a+b)$

$sin^{2}(x) + cos^2{x} = 1, sin^{2}(x) = 1 - cos^{2}(x)$$tan (x) = \frac{sin (x)}{cos(x)} , cot(x) = \frac{cos(x)}{sin(x)},sec (x) = \frac{1}{cos(x)}, cosec (x) = \frac{1}{sin(x)}$