Question

if tan^2A+cot^2A=10÷3, then find the value of tanA+cotA, tanA-cotA.Also find the value of tanA​

Answer 1

$tan²O + Cot²O = \frac{10}{3}$ ----(1)

$tan²O + Cot²O + 2 = \frac{10}{3} + 2$

$(tanO + CotO)² = \frac{10}{3} + \frac{6}{3}$

$(tanO + CotO)² = \frac{16}{3}$

$(<strong>tanO + CotO) = \frac{4}{√3}$ -----(2)

On subtracting 2 from eq.(1)

$tan²O + Cot²O - 2 = \frac{10}{3} - 2$

$(tanO - CotO)² = \frac{10}{3} - \frac{6}{3}$

$(tanO - CotO)³ = \frac{4}{3}$

$<strong>(tanO - CotO) = \frac{2}{√3}</strong>-----(3)$

Now ,

$2tanO = \frac{6}{√3}$

$tanO = \frac{3}{√3}$

$<strong>tanO = √3</strong>$

(O = 60°)