If tan ((απ/4)) = cot ((βπ/4)), then

Question

If tan ((απ/4)) = cot ((βπ/4)), then (A) α+β =0 (B) α+β =2n (C) α+β =2n+1 (D) α+β =2(2n+1), n is an integer

Tardigrade Admin · Accepted Answer

Since, tan ((απ/4)) = cot ((βπ/4)) ⇒ tan ((απ/4)) = tan ((π/2) -(βπ/4)) ⇒ (απ/4) = nπ +(π/2) -(βπ/4) ⇒ (α+β) (π/4) = ((2n+1/2))π ⇒ α +β = 2(2n+1), ∀ n ∈ I

Q. If $tan (\frac{α π}{4}) = cot (\frac{β π}{4})$ , then

$α + β = 0$

$α + β = 2 n$

$α + β = 2 n + 1$

$α + β = 2 (2 n + 1), n$ is an integer

Q. If tan(4απ​)=cot(4βπ​), then

α+β=0

α+β=2n

α+β=2n+1

α+β=2(2n+1),n is an integer

Since, tan(4απ​)=cot(4βπ​) ⇒tan(4απ​)=tan(2π​−4βπ​) ⇒4απ​=nπ+2π​−4βπ​ ⇒(α+β)4π​=(22n+1​)π ⇒α+β=2(2n+1),∀n∈I

Q. If $tan (\frac{α π}{4}) = cot (\frac{β π}{4})$ , then

$α + β = 0$

$α + β = 2 n$

$α + β = 2 n + 1$

$α + β = 2 (2 n + 1), n$ is an integer