Question

If tan 20° = p, prove that $\frac{tan 610\textdegree + tan 700\textdegree}{tan 560\textdegree - tan 470\textdegree}$ = $\frac{1 - p^{2}}{1 + p^{2}}$

Answer 1

tan610° = tan(720° - 110°) = -tan110°
= -tan(90° + 20°) = cot20° = 1/tan20° = 1/p

tan700° = tan(720° - 20°) = -tan20° = -p

tan560° = tan(360° + 200°) = tan200°
= tan(180° + 20°) = tan20° = p

tan470° = tan(360° + 110°) = tan110°
= tan(90° + 20°) = -cot20° = -1/tan20° =- 1/p

now, LHS = (tan610° + tan700°}/(tan560° - tan470°)

= (1/p - p)/(p + 1/p)

= (1 - p²)/(p² + 1)

= (1 - p²)/(1 + p²) = RHS

Answer 2

HELLO DEAR,



GIVEN:- tan60° = p

we know:-
tan700° = tan(720° - 20°) = -tan20° = -p

tan610° = tan(720° - 110°) = -tan110°
= -tan(90° + 20°) = cot20° = 1/tan20° = 1/p

tan560° = tan(360° + 200°) = tan200°
= tan(180° + 20°) = tan20° = p

tan470° = tan(360° + 110°) = tan110°
= tan(90° + 20°) = -cot20° = -1/tan20° =- 1/p

now,

(tan610° + tan700°}/(tan560° - tan470°)

=> (1/p - p)/(p + 1/p)

=> (1 - p²)/(p² + 1)

=> (1 - p²)/(1 + p²)


I HOPE IT'S HELP YOU DEAR,
THANKS