Question

If Ø is an acute angle and tanØ + CotØ= 2 then find the value of tan^7 Ø + Cot^7 Ø.​

Answer 1

Hi there !!

Ø = A ( I am taking A instead of thetha)

It is given that, tanA+ CotA= 2

TanA + 1/tanA = 2 {We now that cotA =1/tanA)

Now, tan²A+ 1=2tanA

Tan²A-2tanA+1=0

(Let tanA be x so it will become quadratic equation)

TanA + 1/tanA = 2

{We now that cotA =1/tanA)

Now,

tan²A+ 1=2tanA

Tan²A-2tanA+1=0(Let tanA be x so it will become quadratic equation)

x²-2x+1=0

x²-x-x+1=0

x(x-1) -1(x-1)

(x-1)²=0 or x=1

Means, tanA =1 Value of tan^7+cot^7

(1)^7+(1/1)^7= 2

Thankyou :)

Answer 2

Given,

tanØ + CotØ = 2

SinØ/CosØ + CosØ/SinØ = 2

(Sin^2Ø + Cos^2Ø)/SinØ.CosØ = 2

=> 1/SinØ.CosØ = 2 ( sin^2Ø+ Cos^2Ø=1)

=> 1/2SinØ. CosØ = 2/2 (Dividing both sides by 2).

=> 1/Sin2Ø = 1 ( 2SinØ. CosØ = Sin2Ø)

=> Sin2Ø = Sin 90°

=> 2Ø = 90°

=> Ø = 45°

Again,

Expression :-

=> tan^7Ø + Cot^7Ø

=> tan^7 45° + Cot^7 45°

=> 1^7 + 1^7 ( tan 45° = Cot 45° = 1)

=> 2