If beginpmatrix1 - tan θ tan θ 1 endpmatrix beginpmatrix1 tan θ - tan θ 1 endpmatrix-1= beginpmatrixa -b b a endpmatrix, then

Question

If beginpmatrix1 - tan θ tan θ 1 endpmatrix beginpmatrix1 tan θ - tan θ 1 endpmatrix-1= beginpmatrixa -b b a endpmatrix, then (A) a=b=1 (B) a= cos 2 θ, b= sin 2 θ (C) a= sin 2 θ, b= cos 2 θ (D) a=1, b= sin 2 θ

Tardigrade Admin · Accepted Answer

beginpmatrixa -b b a endpmatrix = beginpmatrix 1 - tan θ tan θ 1 endpmatrix beginpmatrix cos 2 θ - sin θ cos θ sin θ cos θ cos 2 θ endpmatrix = beginpmatrix cos 2 θ - sin 2 θ sin 2 θ cos 2 θ endpmatrix ⇒ a= cos 2 θ, b= sin 2 θ

Q. If $(1 tan θ - tan θ 1) (1 - tan θ tan θ 1)^{- 1} = (a b - b a)$ , then

$a = b = 1$

$a = cos 2 θ, b = sin 2 θ$

$a = sin 2 θ, b = cos 2 θ$

$a = 1, b = sin 2 θ$

$(a b - b a) = (1 tan θ - tan θ 1) (cos^{2} θ sin θ cos θ - sin θ cos θ cos^{2} θ)$
$= (cos 2 θ sin 2 θ - sin 2 θ cos 2 θ)$
$\Rightarrow a = cos 2 θ, b = sin 2 θ$

Q. If (1tanθ​−tanθ1​)(1−tanθ​tanθ1​)−1=(ab​−ba​), then

a=b=1

a=cos2θ,b=sin2θ

a=sin2θ,b=cos2θ

a=1,b=sin2θ