Question

Prove that (tan theta-cot theta) / (sin theta * cos theta)=(tan² theta-cot² theta)

Answer 1

this is the answer in the attached file

Answer 2

$\frac{tan@-cot@}{sin@.cos@}= tan^{2}@- cot^{2}@$
LHS= $\frac{tan@-cot@}{sin@.cos@} = \frac{tan@- \frac{1}{tan@} }{sin@.cos@}$
$= \frac{ tan^{2}@-1 }{tan@.sin@.cos@} = \frac{ tan^{2}@-1 }{tan@ \frac{sin@}{cos@} .cos@.cos@}$
$= \frac{( tan^{2}@-1) sec^{2}@ }{ tan^{2}@ } = \frac{( tan^{2}@-1)( tan^{2}@+1) }{ tan^{2}@ }$
$= \frac{ tan^{4}@-1 }{ tan^{2}@ } = \frac{ tan^{4}@ }{ tan^{2}@ } - \frac{1}{ tan^{2}@ }$
$= tan^{2}@- cot^{2}@ = R.H.s$