Skip to content
NCERT Exemplar · Q41

Q.If cos⁡2θ=0\cos 2\theta = 0, then ∣0cos⁡θsin⁡θcos⁡θsin⁡θ0sin⁡θ0cos⁡θ∣2=\begin{vmatrix} 0 & \cos\theta & \sin\theta \\ \cos\theta & \sin\theta & 0 \\ \sin\theta & 0 & \cos\theta \end{vmatrix}^2 = ________ .

Dnh Dd CbseShort· 1mImportance★★★★★
83% · 121/146 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The determinant equals −(cos⁡3θ+sin⁡3θ)-(\cos^3\theta+\sin^3\theta); with cos⁡2θ=0\cos 2\theta = 0 its square is 12\tfrac12.

Set up

Write a=cos⁡θa=\cos\theta and b=sin⁡θb=\sin\theta, so the determinant is

Δ=∣0abab0b0a∣.\Delta = \begin{vmatrix} 0 & a & b \\ a & b & 0 \\ b & 0 & a \end{vmatrix}.

Expand along the first row

Δ=0⋅(… )−a∣a0ba∣+b∣abb0∣=−a(a2)+b(−b2)=−(a3+b3).\Delta = 0\cdot(\dots) - a\begin{vmatrix} a & 0 \\ b & a \end{vmatrix} + b\begin{vmatrix} a & b \\ b & 0 \end{vmatrix} = -a(a^2) + b(-b^2) = -(a^3+b^3).

So Δ=−(cos⁡3θ+sin⁡3θ)\Delta = -(\cos^3\theta+\sin^3\theta) and Δ2=(cos⁡3θ+sin⁡3θ)2\Delta^2 = (\cos^3\theta+\sin^3\theta)^2.

Use the condition

cos⁡2θ=0⇒cos⁡2θ=sin⁡2θ=12\cos 2\theta = 0 \Rightarrow \cos^2\theta = \sin^2\theta = \tfrac12, hence cos⁡θsin⁡θ=±12\cos\theta\sin\theta = \pm\tfrac12. Expanding the square,

Δ2=cos⁡6θ+sin⁡6θ+2cos⁡3θsin⁡3θ=(1−3cos⁡2θsin⁡2θ)+2cos⁡3θsin⁡3θ.\Delta^2 = \cos^6\theta+\sin^6\theta + 2\cos^3\theta\sin^3\theta = \big(1 - 3\cos^2\theta\sin^2\theta\big) + 2\cos^3\theta\sin^3\theta. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.