Question

prove
3-4sin²∅÷cos²∅=3-tan²∅

Answer 1

LHS = ( 3 - 4sin²∅) ÷cos²∅

= 3/cos²∅ - (4sin²∅/cos²∅ )

= 3sec²∅ - 4tan²∅

= 3sec²∅ - 3tan²∅ - tan²∅

= 3(sec²∅ - tan²∅ )- tan²∅

$\boxed { \pink { sec^{2} A - tan^{2} A = 1 }}$

= 3 - tan²∅

= RHS

•••♪