Question

If tan20=p, then prove that tan 6100+tan 7000 tan 5600−tan 4700 = 1−p 2 1+p

Answer 1

Answer:

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Answer 2

Answer:

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

Step-by-step explanation:

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