Question

prove that tan theta+tan[90-theta]=sec theta*sec[90-theta]

Answer 1

$\mathfrak{ HELLO\:\: DEAR,}$

$TAN\theta + TAN(90 - \theta) \\ \\ \mathit{TAN\theta + COT\theta} \\ \\ \mathit{\frac{SIN\theta}{COS\theta} + \frac{COS\theta}{SIN\theta}}$

$\mathit{= \frac{SIN^2\theta + COS^2\theta}{SIN\theta*COS\theta}}$


$\mathit{= \frac{1}{SIN\theta*COS\theta}} \\ \\ \mathit{= COSEC\theta*SEC\theta} \\ \\ \mathit{= SEC(90 - \theta)*SEC\theta}$


$\: \: \: \: \: \: \: \: \: \: \: \to \to \to \: \: \: \therefore \mathit{\boxed{SEC(90 - \theta) = COSEC\theta}}$


$\underline \mathit{{I \:\: HOPE\:\: IT'S\:\: YOU\:\: DEAR\:\: THANKS}}$

Answer 2

Hope it helps u.....