KCET · Maths · Determinants
The value of \(\left|\begin{array}{lll}\sin ^2 14^{\circ} & \sin ^2 66^{\circ} & \tan 135^{\circ} \\ \sin ^2 66^{\circ} & \tan 135^{\circ} & \sin ^2 14^{\circ} \\ \tan 135^{\circ} & \sin ^2 14^{\circ} & \sin ^2 66^{\circ}\end{array}\right|\) is
- A \(0\)
- B \(1\)
- C \(2\)
- D \(-1\)
Answer & Solution
Correct Answer
(A) \(0\)
Step-by-step Solution
Detailed explanation
Here,
\(\left|\begin{array}{ccc}\sin ^2 14^{\circ} & \sin ^2 66^{\circ} & \tan 135^{\circ} \\ \sin ^2 66^{\circ} & \tan 135^{\circ} & \sin ^2 14^{\circ} \\ \tan 135^{\circ} & \sin ^2 14^{\circ} & \sin ^2 66^{\circ}\end{array}\right|\)
Applying \(C_1 \rightarrow C_1+C_2+C_3\)
\(=\left|\begin{array}{lll}\sin ^2 14^{\circ}+\sin ^2 66^{\circ}+\tan 135^{\circ} & \sin ^2 66^{\circ} \tan 135^{\circ} \\ \sin ^2 14^{\circ}+\sin ^2 66^{\circ}+\tan 135^{\circ} & \tan 135^{\circ} \sin ^2 14^{\circ} \\ \sin ^2 14^{\circ}+\sin ^2 66^{\circ}+\tan 135^{\circ} & \sin ^2 14^{\circ} \sin ^2 66^{\circ}\end{array}\right|\)
\(\begin{aligned} & =\sin ^2 14^{\circ}+\sin ^2 66^{\circ}+\tan 135^{\circ} \\ & \qquad\left|\begin{array}{lll}1 & \sin ^2 66^{\circ} & \tan 135^{\circ} \\ 1 & \tan 135^{\circ} & \sin ^2 14^{\circ} \\ 1 & \sin ^2 14^{\circ} & \sin ^2 66^{\circ}\end{array}\right|\end{aligned}\)
\(=\sin ^2 14^{\circ}+\sin ^2 66^{\circ}-1\left|\begin{array}{ccc}1 & \sin ^2 66 & -1 \\ 1 & -1 & \sin ^2 14^{\circ} \\ 1 & \sin ^2 14^{\circ} & \sin ^2 66^{\circ}\end{array}\right|\)
Applying \(R_1 \rightarrow R_1-R_2, R_2 \rightarrow R_2-R_3\)
\(\begin{aligned} & =\cos ^2 66^{\circ}+\sin ^2 66^{\circ}-1 \\ & \left|\begin{array}{ccc}0 & 1+\sin ^2 66^{\circ} & -1-\sin ^2 14^{\circ} \\ 0 & -1-\sin ^2 14^{\circ} & \sin ^2 14^{\circ}-\sin ^2 66^{\circ} \\ 1 & \sin ^2 14^{\circ} & \sin ^2 66^{\circ}\end{array}\right| \\ & \left(\cos ^2 66^{\circ}+\sin ^2 66^{\circ}-1\right)=0\end{aligned}\)
\(\left|\begin{array}{ccc}\sin ^2 14^{\circ} & \sin ^2 66^{\circ} & \tan 135^{\circ} \\ \sin ^2 66^{\circ} & \tan 135^{\circ} & \sin ^2 14^{\circ} \\ \tan 135^{\circ} & \sin ^2 14^{\circ} & \sin ^2 66^{\circ}\end{array}\right|\)
Applying \(C_1 \rightarrow C_1+C_2+C_3\)
\(=\left|\begin{array}{lll}\sin ^2 14^{\circ}+\sin ^2 66^{\circ}+\tan 135^{\circ} & \sin ^2 66^{\circ} \tan 135^{\circ} \\ \sin ^2 14^{\circ}+\sin ^2 66^{\circ}+\tan 135^{\circ} & \tan 135^{\circ} \sin ^2 14^{\circ} \\ \sin ^2 14^{\circ}+\sin ^2 66^{\circ}+\tan 135^{\circ} & \sin ^2 14^{\circ} \sin ^2 66^{\circ}\end{array}\right|\)
\(\begin{aligned} & =\sin ^2 14^{\circ}+\sin ^2 66^{\circ}+\tan 135^{\circ} \\ & \qquad\left|\begin{array}{lll}1 & \sin ^2 66^{\circ} & \tan 135^{\circ} \\ 1 & \tan 135^{\circ} & \sin ^2 14^{\circ} \\ 1 & \sin ^2 14^{\circ} & \sin ^2 66^{\circ}\end{array}\right|\end{aligned}\)
\(=\sin ^2 14^{\circ}+\sin ^2 66^{\circ}-1\left|\begin{array}{ccc}1 & \sin ^2 66 & -1 \\ 1 & -1 & \sin ^2 14^{\circ} \\ 1 & \sin ^2 14^{\circ} & \sin ^2 66^{\circ}\end{array}\right|\)
Applying \(R_1 \rightarrow R_1-R_2, R_2 \rightarrow R_2-R_3\)
\(\begin{aligned} & =\cos ^2 66^{\circ}+\sin ^2 66^{\circ}-1 \\ & \left|\begin{array}{ccc}0 & 1+\sin ^2 66^{\circ} & -1-\sin ^2 14^{\circ} \\ 0 & -1-\sin ^2 14^{\circ} & \sin ^2 14^{\circ}-\sin ^2 66^{\circ} \\ 1 & \sin ^2 14^{\circ} & \sin ^2 66^{\circ}\end{array}\right| \\ & \left(\cos ^2 66^{\circ}+\sin ^2 66^{\circ}-1\right)=0\end{aligned}\)
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