Question: If tanA + cotA= 2 then tan²A - cot²A = ?

Posted by: Aditya K. on 10.01.2020

Given: tanA + cotA = 2

To find: tan2A- cot2A

Sol:     (tanA - cotA)2 =(tanA + cotA)2 - 4tanAcotA

= 4 - 4

(tanA - cotA)2 = 0

tanA - cotA   = 0

=>    tan2A - cot2A  = (tanA + cotA) (tanA - cotA)

= 2 X 0

=0   (Ans.)

This is a solution

Tan^A-Cot^A=(TanA+CotA)(TanA-CotA)=2.0=0

TanA+CotA =2

(TanA-CotA)^2= (TanA+CotA)^2-4TanA.CotA=2^-4.1  (TanA.CotA=1) =2^2-4=4-4=0

(TanA-CotA)^2=0

TanA-CotA=0

Kindly see the below attachment

TanA=cotA. And A=45 Thus reuired answer =0
TanA^2-2tanA+1=0 (tanA-1)^2=0 tanA-1=0 tanA=1 So,tanA-cotA=tanA-1/tanA=1-1=0
TanA +CotA= 2. (tanA-cotA) sqare= (tanA+cotA) sqare -4tanA.cotA =4-4 (as tanA.cotA=1) =0 So tan(sqare)A - cot(sqare)A = (tanA+cotA).(tanA-cotA) = 2*0 = 0 Ans.

Given, tan A + cot A = 2

thus, tan A + 1/tan A = 2

or (tan A)^2 + 1 = 2tan A

or (tan A)^2 - 2tan A + 1 = 0

or (tan A)^2 - tan A - tan A + 1 = 0

or tan A (tan A - 1) - 1(tan A - 1) = 0

or, (tan A - 1)^2 = 0

or tan A - 1 = 0

or tan A = 1

thus, cot A = 1

hence, (tan A)^2 - (cot A)^2 = 1 - 1 = 0

tan A +cotA = 2 means Tan A= Cot A =1 .

If tanA + cotA = 2, then tanA + 1/tanA =2. This equation is of the x +1/x = 2. Now for all values of x, x + 1/x is always greater than equal to 2. It is equal to 2 when x =1. Hence tan A = 1, implying A =45 degrees. Thus tan^2 A - cot^2 A is actually 0, since both tan 45 and cot 45 =1

Given, tan A + cot A = 2

Now, tan^2 A - cot^2 A = (tan A - cot A)(tan A + cot A)

= 2(tan A - cot A) ........ (i)

But, tan A - cot A = [(tan A - cot A)^2]^1/2

= [(tan A + cot A)^2 - 4tan A cot A]^1/2

= [4 - 4]

= 0 ......... (ii)

Substituting (ii) in (i), we get,

tan^2 A - cot^2 A = (tan A - cot A)(tan A + cot A)

= 2 x 0

= 0

Hence, tan^2 A - cot^2 A = 0

Given Tan A + Cot A =1 , Usually Tan 45 = Cot 45 =1 (45 degrees) . With which Its very Obvious that A = 45 degrees.

This implies Tan^2 A - Cot^2 A = 1 -1 = 0

tan A+ cotA=2

WE KNOW THAT, TAN A.COT A=1

NOW,

(TAN A- COT A)^2=(TAN A+ COT A)^2 -4TAN A.COT A

(TAN A-COT A)^2=4-4

TAN A- COT A=0

TAN^2 A- COT^2 A=(TAN A+ COT A)(TAN A- COT A)

=2*0=0

Using the formula, (a-b)^2 = (a+b)^2 - 4ab, here we get,

(tanA-cotA)^2 = 2^2 - 4x1 = 4 - 4 = 0.

So, (tanA - cot A) = 0.

Now, tan^2 A - cot^2 A = (tanA + cotA).(tanA - cotA) = 2x0 = 0 .

Ans. 0

pl see the attached file for answer

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