Question

Prove that:

`cot theta tan (90^\circ - theta) - sec (90^\circ - theta) "cosec"  theta + sqrt 3 tan 12^\circ tan 60^\circ tan 78^circ` = 2

Answer 1

LHS = `cot theta tan (90^\circ - theta) - sec (90^\circ - theta) "cosec"  theta + sqrt 3 tan 12^\circ tan 60^\circ tan 78^circ`

= `cot theta * cot theta - "cosec" theta * "cosec" theta + sqrt 3 tan 12^\circ * sqrt 3 tan 78^\circ`

= `cot^2 theta - "cosec'^2 theta + 3 tan 12^\circ tan 78^\circ`

= `cot^2 theta - (1 + cot^2 theta) + 3 tan 12^\circ cot 12^\circ`

= −1 + 3(1)

= 2

= RHS