tanA+cotA=2
tanA-cotA=?
Answers
Answer:
Given:
tanA+cotA=2
tanA+(1/tanA)=2. (cotA=1/tanA)
(tanA)^2 + 1 =2tanA
(tanA)^2 - 2tanA + 1 = 0
(tanA - 1)^2 = 0.
so tanA=1
if tanA=1, then cotA=1
now according to question;
tan^7(A) +cot^7(A)=?
(1)^7 + (1)^7=2
hence, tan^7(A) +cot^7(A)=2
If tan + cot=2, then what is the value of tan^3+7cot^3?
If tanθ+cotθ=2 then what is the value of tan5θ+cot5θ ?
How do I prove tan(π/7)⋅tan(2π/7)⋅tan(3π/7)=7–√ ?
If tanA+cotA=3, then what is tan^8A+cot^8A?
If tanθ+cotθ=2 then what is the value of tan7θ+cot7θ ?
Prerequisites:-
tanA=1cotA
(x−y)2=x2−2.x.y+y2
Given,
tanA+cotA=2
From Prerequisite 1
tanA+1tanA=2
tan2A+1tanA=2
tan2A+1=2.tanA
tan2A−2.tanA+12=0
From Prerequisite 2
(tanA−1)2=0
tanA−1=0
tanA=1
cotA=1
So ,
tan7A+cot7A
(tanA)7+(cotA)7A
Putting value of tanA and cotA
1+1=2
So answer is 2
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Write cotA=1/TanA and take LCM. you will get a quadratic equation
Tan^2 (A) -2 Tan A +1 =0 which is (Tan A -1 )^2 =0
So tan A has only one value which is 1
Then Tan ^ 7 (A) + Cot^7 (A) = 1+1 =2
If tanA+cotA=4, then what is the value of tan^6A+cot^6A?
If tanA+cotA=5, what is tan^2A+cot^2A?
Is tan^2(x)+cot^2(x)=2?
What is the value of sin x if (tan x + cot x) / (tan x - cot x) = 2?
How can I prove, tan A/1-cot A + cot A/1-tan A = 1+ tan A + cot A?
tan A+ cot A= 2
tan A+ 1/tan A= 2
tan^2 A+1= 2tan A
tan^2 A- tan A- tan A+1= 0
tan A(tan A-1)- 1(tan A-1)= 0
tan A= 1
Therefore, A equals 45°.
Therefore,
(tan 45°)^7 + (cot 45°)^7= 1+1= 2
So, the final answer is 2.
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tanA+cotA=2
tanA+1tanA=2
tan2A+1tanA=2
tan2A+1=2tanA
let tan A be x
x2+1=2x
x2−2x+1=0
(x−1)2=0
x=1
=>tanA=x=1
=>tanA=cotA=1
=>tan7A+cot7A=1 ….. req ans
TanA=1/cotA
Let us assume tanA=x where X belongs to (-infinite, +infinite) except 0
Now this given function will be like
x+1/x=2
As (x+1/x) has range [-2,infinity) or[2,infinity)
So equality hold only x=1
So tanA=1
TanA^7+cotA^7=2
tanA+cotA=2
(tanA+cotA)²=4
(tanA-cotA)²+4=4
(tanA-cotA)²=0
tanA-cotA=0