If θ=π / 4 n, then the value of tan θ tan 2 θ ⋅s tan (2 n-2) θ tan (2 n-1) θ is

Question

If θ=π / 4 n, then the value of tan θ tan 2 θ ⋅s tan (2 n-2) θ tan (2 n-1) θ is (A) -1 (B) 1 (C) 0 (D) 2

Tardigrade Admin · Accepted Answer

2 n θ=π / 2. Thus, θ and (2 n-1) θ=(π / 2)-θ ; 2 θ and (2 n-2) θ=(π / 2)-2 θ, ⋅s form complementary angles A and B so that tan A tan B= tan A cot A=1 for each pair.

Q. If $θ = π /4 n$ , then the value of $tan θ tan 2 θ \dots tan (2 n - 2)$ $θ tan (2 n - 1) θ$ is

-1

1

0

2

$2 n θ = π /2$ . Thus,
$θ$ and $(2 n - 1) θ = (π /2) - θ; 2 θ$ and $(2 n - 2) θ = (π /2) - 2 θ, \dots$
form complementary angles $A$ and $B$ so that $tan A tan B = tan A cot A = 1$ for each pair.

Q. If θ=π/4n, then the value of tanθtan2θ⋯tan(2n−2) θtan(2n−1)θ is

-1

1

0

2

2nθ=π/2. Thus, θ and (2n−1)θ=(π/2)−θ;2θ and (2n−2)θ=(π/2)−2θ,⋯ form complementary angles A and B so that tanAtanB=tanAcotA=1 for each pair.

Q. If $θ = π /4 n$ , then the value of $tan θ tan 2 θ \dots tan (2 n - 2)$ $θ tan (2 n - 1) θ$ is

$2 n θ = π /2$ . Thus,
$θ$ and $(2 n - 1) θ = (π /2) - θ; 2 θ$ and $(2 n - 2) θ = (π /2) - 2 θ, \dots$
form complementary angles $A$ and $B$ so that $tan A tan B = tan A cot A = 1$ for each pair.