If f(θ )=| beginmatrix cos 2θ cos θ sin θ - sin θ cos θ sin 2θ cos θ sin θ - cos θ 0 endmatrix | Then, for all θ

Question

If f(θ )=| beginmatrix cos 2θ cos θ sin θ - sin θ cos θ sin 2θ cos θ sin θ - cos θ 0 endmatrix | Then, for all θ  (A)  f(θ )=0  (B)  f(θ )=1  (C)  f(θ )=-1  (D) None of these

Tardigrade Admin · Accepted Answer

Given, f(θ )=| beginmatrix cos 2θ cos θ sin θ - sin θ cos θ sin θ sin 2θ cos θ sin θ - cos θ 0 endmatrix | = cos 2θ (0+ cos 2θ )- cos θ . sin θ (0- sin θ . cos θ ) - sin θ (- cos 2θ . sin θ - sin 3θ ) = cos 4θ +2 sin 2θ . cos 2θ + sin 4θ =( cos 2θ + sin 2θ )2 =1 ∴ For all θ ,f(θ )=1

Q. If $f (θ) = ∣ ∣ cos^{2} θ cos θ sin θ cos θ sin θ sin^{2} θ - cos θ - sin θ cos θ 0 ∣ ∣$ Then, for all $θ$

$f (θ) = 0$

$f (θ) = 1$

$f (θ) = - 1$

None of these

Q. If f(θ)=∣∣​cos2θcosθsinθ​cosθsinθsin2θ−cosθ​−sinθcosθ0​∣∣​ Then, for all θ

f(θ)=0

f(θ)=1

f(θ)=−1

None of these

Given, f(θ)=∣∣​cos2θcosθsinθsinθ​cosθsinθsin2θ−cosθ​−sinθcosθ0​∣∣​ =cos2θ(0+cos2θ)−cosθ .sinθ(0−sinθ.cosθ) −sinθ(−cos2θ.sinθ−sin3θ) =cos4θ+2sin2θ.cos2θ+sin4θ =(cos2θ+sin2θ)2 =1 ∴ For all θ,f(θ)=1

Q. If $f (θ) = ∣ ∣ cos^{2} θ cos θ sin θ cos θ sin θ sin^{2} θ - cos θ - sin θ cos θ 0 ∣ ∣$ Then, for all $θ$

$f (θ) = 0$

$f (θ) = 1$

$f (θ) = - 1$