Evaluate the determinant - cos- -sin- sin- cos-

cosθ #160; amp; sinθ #160; sinθ #160; amp; cosθ #160; #160; Expanding the determinant, we get cos^2θ ( sin ^2θ) cos^2θ+sin ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Determinant

Evaluate the ...

Question

Evaluate the determinant :
$∣ ∣ ∣ \begin{matrix} cos θ & - sin θ sin θ & cos θ \end{matrix} ∣ ∣ ∣$

Open in App

Solution

$∣ ∣ ∣ \begin{matrix} cos θ & - sin θ sin θ & cos θ \end{matrix} ∣ ∣ ∣$

Expanding the determinant, we get

${cos}^{2} θ - (- {sin}^{2} θ)$

${cos}^{2} θ + {sin}^{2} θ$

$= 1$ .............. $∵ {cos}^{2} θ + {sin}^{2} θ = 1$

Suggest Corrections

Similar questions

Evaluate the determinant
$(i)$ $∣ ∣ ∣ \begin{matrix} cos θ & - sin θ sin θ & cos θ \end{matrix} ∣ ∣ ∣$
$(i i)$ $∣ ∣ ∣ \begin{matrix} x^{2} - x + 1 & x - 1 x + 1 & x + 1 \end{matrix} ∣ ∣ ∣$

Q. The value of the determinat

(\begin{matrix} cos θ & - sin θ sin θ & cos θ \end{matrix} ∣ ∣ ∣

is .

Q. If

A = [\begin{matrix} c o s θ & - s i n θ s i n θ & c o s θ \end{matrix}]

, then

A^{T} + A = I_{2}

, if

Q. If

I = I = [\begin{matrix} 10 \end{matrix} \begin{matrix} 01 \end{matrix}], j = [\begin{matrix} 0 - 1 \end{matrix} \begin{matrix} 10 \end{matrix}] a n d B = [\begin{matrix} c o s θ - s i n θ \end{matrix} \begin{matrix} s i n θ c o s θ \end{matrix}],

then B =

Q. Evaluate

tan θ . sin θ + cos θ = sec θ

Join BYJU'S Learning Program

Evaluate the determinant :∣∣∣cosθ−sinθsinθcosθ∣∣∣

∣∣∣cosθ−sinθsinθcosθ∣∣∣Expanding the determinant, we getcos2θ−(−sin2θ)cos2θ+sin2θ=1..............∵cos2θ+sin2θ=1

Evaluate the determinant :
$∣ ∣ ∣ \begin{matrix} cos θ & - sin θ sin θ & cos θ \end{matrix} ∣ ∣ ∣$

$∣ ∣ ∣ \begin{matrix} cos θ & - sin θ sin θ & cos θ \end{matrix} ∣ ∣ ∣$

Expanding the determinant, we get

${cos}^{2} θ - (- {sin}^{2} θ)$

${cos}^{2} θ + {sin}^{2} θ$

$= 1$ .............. $∵ {cos}^{2} θ + {sin}^{2} θ = 1$