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Question : 8 of 160
Marks:
+1,
-0
Solution:
Let Δ =
|| cos‌2‌x | sin2x | cos‌2‌x |
| sin2x | cos‌2‌x | cos2x |
| cos‌2‌x | cos2x | cos‌2‌x‌ |
| On expanding along
R1.
Δ=cos‌2‌x[cos22x−cos4x]−sin2x [sin2x‌cos‌2‌x−cos2x.cos‌2‌x]+cos‌2‌x[sin2xcos2x−cos22x]
Δ=cos‌2‌x[(2cos2x−1)2−cos4x]−sin2x[cos‌2‌x(sin2x−cos2x)]+cos‌2‌x[cos2x−cos4x−4cos4x−1+4cos2x]
=(2cos2x−1)[3cos4x+1+5cos2x−5cos4x−1]+(1−cos2x)[(2cos2x−1)2]
⇒(2cos2x−1)[5cos2x−2cos4x]+(2cos2x−1)[2cos2x−1−2cos3x+cos2x]
⇒(2cos2x−1)[5cos2x−2cos4x+3cos2x−2cos3x−1]
⇒
(2cos2x−1)[8cos2x−2cos4x−2cos3x−1] Hence, constant term will be
(−1)×(−1)=1
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