Question

in a triangle ABC if angle C = 90° prove that cosec²A - tan²b = 1

Answer 1

hey mate
here's the solution

Answer 2

Let this be your right angled triangle,
where angle C = 90° .


Now,
${ \csc(a) }^{2} = {( \frac{ab}{cb} )}^{2} = \frac{{ab}^{2} }{ {cb}^{2} }$




And

${ \tan(b) }^{2} = \frac{ {ac}^{2} }{ {cb}^{2} }$



Then,

${ \csc(a) }^{2} - { \tan(b) }^{2} \\ = \frac{ {ab}^{2} }{ {cb}^{2} } - \frac{ {ac}^{2} }{ {cb}^{2} } \\ = \frac{ {ab}^{2} - {ac}^{2} }{ {cb}^{2} }$


By Pythagoras theorem the denominator ac squared.

Therefore,
it becomes

$\frac{ {cb}^{2} }{ {cb}^{2} } = 1 = rhs$

Hence proved.