Question

Tan^5x+cot^5x=2525 then tan^2x+cot^2x=?

Answer 1

it is given that, tan^5x + cot^5x = 2525 and then we have to find out the value of tan²x + cot²x

tan^5x + cot^5x = 25

⇒ (tan²x + cot²x)(tan³x + cot³x) - tan²x.cot³x - tan³x.cot²x = 25

⇒(tan²x + cot²x)(tan³x + cot³x) - (tanx + cotx) = 2525

we know, tan²x + cot²x = (tanx + cotx)² - 2

(tan³x + cot³x) = (tanx + cotx)³ - 3(tanx + cotx)

so, (tan²x + cot²x)(tan³x + cot³x) - (tanx + cotx) = { (tanx + cotx)² - 2}{(tanx + cotx)³ - 3(tanx + cotx)} - (tanx + cotx) = 2525

Let (tanx + cotx) = t

⇒(t² - 2)(t³ - 3t) - t = 2525

⇒t^5 - 3t³ - 2t³ + 6t - t = 2525

⇒t^5 - 5t³ + 5t - 2525 = 0

if t = 5, (5)^5 - 5(5)³ + 5(5) - 2525 = 0

so, (tanx + cotx) = 5

now, (tan²x + cot²x) = (tan + cotx)² - 2

= (5)² - 2 = 23

hence, tan²x + cot²x = 23