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Trigonometry question?

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If tan theta + cot theta = 2, show tan^10 theta + cot^10 theta =2. thanks in advance.
asked Dec 21, 2012 in TRIGONOMETRY by dozey Mentor

3 Answers

0 votes

Let tan theta = x

cot theta= 1/x

tan theta + cot theta= x+ 1/x = 2

x^2 + 1 = 2 x

or (x^2-2x+1) = 0

(x-1) ^2 = 0

x = 1

tan theta = cot theta = 1

tan ^10 theta + cot ^10 theta

=( 1)^10+(1)^10

= 2

tan ^10 theta + cot ^10 theta = 2

answered Dec 25, 2012 by ashokavf Scholar
0 votes

cosθ

Let tan θ = x

cot θ= 1/x

tan θ + cot θ= x+ 1/x = 2

x2 + 1 = 2 x

or (x2-2x+1) = 0

(x-1)2 = 0

(x-1)(x-1) = 0

x-1 = 0 ,x-1 = 0

x = 1

tan θ = cot θ = 1

tan
10 θ + cot 10 θ

=( 1)10+(1)10

= 2

tan 10 θ + cot 10 θ = 2

There fore          tan 10 θ + cot 10 θ = 2

answered Jan 9, 2013 by richardson Scholar
0 votes

tanθ + cotθ = 2

tanθ + 1/tanθ = 2

(tanθ×tanθ + 1) / tanθ = 2

tan2 θ + 1 = 2 tanθ

tan2 θ - 2 tanθ + 1 = 0               [ This is in the form (a - b)2 which is equal to a2- 2ab+b2 ]

(tanθ - 1)2 =0

(tanθ - 1)(tanθ - 1) = 0

tanθ - 1 = 0

tanθ = 1

=> cotθ = 1/tanθ = 1/1 = 1

Therefore tan10 θ + cot10 θ = (tanθ)2 +(cotθ)2 = 110 + 110 = 1+1 = 2.

answered May 21, 2014 by joly Scholar

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