Posted by **aman** on Wednesday, June 13, 2012 at 7:08am.

if tanx+cotx=2 then the value of tan^5x + cot^10x is

- arithmetic -
**Reiny**, Wednesday, June 13, 2012 at 8:55am
If tanx + cotx = 2

sinx/cosx + cosx/sinx = 2

(sin^2 x + cos^2 x)/(sinxcosx) = 2

1/(sinxcosx) = 2

2sinxcosx= 1

sin 2x = 1

2x = 90° for 0 ≤ x ≤ 180°

x = 45° or π/4 radians

we know tan 45° = 1 and cot 45° = 1

then tan^5 x + cot^10 x

= 1^5 + 1^10

= 2

There are other solutions for x

they are x + k(180°) , where k is an integer

but all such angles will fall in either quadrant I or III and both the tangent and cotangent would be +1

so tan^5 x + cot^2 x = 2

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