The value of the determinat cos-sin-sin-cos- is

The correct option is D 1cos nbsp;θ sin nbsp;θsin nbsp;θcos nbsp;θ nbsp;= nbsp;cos nbsp;θcos nbsp;θ nbsp;sin nbsp;θ sin nbsp;θ= nbsp;co ...

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Byju's Answer

Standard XII

Mathematics

Evaluation of a Determinant

The value of ...

Question

The value of the determinat $(\begin{matrix} cos θ & - sin θ sin θ & cos θ \end{matrix} ∣ ∣ ∣$ is .

2 {sin}^{2} θ

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2 {cos}^{2} θ

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0

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1

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Solution

The correct option is D $1$
$(\begin{matrix} cos θ & - sin θ sin θ & cos θ \end{matrix} ∣ ∣ ∣ = cos θ (cos θ) - sin θ (- sin θ) = {cos}^{2} θ + {sin}^{2} θ = 1$

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Similar questions

Q. Find the value of determinant.
(i)

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

(ii)

∣ ∣ ∣ \begin{matrix} x^{2} - x + 1 & x - 1 x + 1 & x + 1 \end{matrix} ∣ ∣ ∣

Q. Evaluate the determinant :

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

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

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. 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 =

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The value of the determinat (cos θ−sin θsin θcos θ∣∣∣ is .

The correct option is D 1(cos θ−sin θsin θcos θ∣∣∣ = cos θ(cos θ)− sin θ(−sin θ)= cos2θ+sin2θ = 1

The value of the determinat $(\begin{matrix} cos θ & - sin θ sin θ & cos θ \end{matrix} ∣ ∣ ∣$ is .

The correct option is D $1$
$(\begin{matrix} cos θ & - sin θ sin θ & cos θ \end{matrix} ∣ ∣ ∣ = cos θ (cos θ) - sin θ (- sin θ) = {cos}^{2} θ + {sin}^{2} θ = 1$