Question

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if tan A + cot A =2 show that tan ^2 A + cot ^2 A =2

Answer 1

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tanA=1cotAtanA=1cotA(x−y)2=x2−2.x.y+y2(x−y)2=x2−2.x.y+y2

Given,

tanA+cotA=2tanA+cotA=2

From Prerequisite 1

tanA+1tanA=2tanA+1tanA=2

tan2A+1tanA=2tan2A+1tanA=2

tan2A+1=2.tanAtan2A+1=2.tanA

tan2A−2.tanA+12=0tan2A−2.tanA+12=0

From Prerequisite 2

(tanA−1)2=0(tanA−1)2=0

tanA−1=0tanA−1=0

tanA=1tanA=1

cotA=1cotA=1

So ,

tan7A+cot7Atan7A+cot7A

(tanA)7+(cotA)7A(tanA)7+(cotA)7A

Putting value of tanAtanA and cotAcotA

1+1=2

So answer is 2

Answer 2

tana + cota =2

take square both sides

(tana + cota)² = 2²

tan²a + cot²a +2tana.cota =4

we Know,

tana =1/cota

so,

tan² a + cot²a +2 = 4

tan²a + cot² a = 4 -2 = 2