Question

prove that (tan³thita-1)/(tan thita-1)= sec²thita+tan thita​

Answer 1

Step-by-step explanation:

tan³∅ -1

tan∅-1

multiplying numerator and dominator with tan∅+1

we have,

(tan³∅-1) (tan∅+1)

(tan∅ -1) (tan∅+1)

= tan⁴∅+tan³∅-tan∅-1

tan²∅-1²

= ( tan²∅)²-1 +tan∅(tan²∅-1)

tan²∅-1

= ( tan²∅-1)(tan²∅+1) +tan∅(tan²∅-1)

tan²∅-1

=(tan²∅-1)(tan²∅+1+tan∅)

tan²∅-1

= tan²∅+1+tan∅

=sec²∅+tan∅

= R.H.S