arithmetic
posted by aman .
if tanx+cotx=2 then the value of tan^5x + cot^10x is

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