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Question : 5 of 150
Marks:
+1,
-0
Solution:
Let A =
|| 1+sin2θ | sin2θ | sin2θ |
| cos2θ | 1+cos2θ | cos2θ |
| 4‌sin‌4‌θ | 4‌sin‌4‌θ | 1+4‌sin‌4‌θ |
|=0 Applying
R1→R1+R2 Applying
C1→C1−2C3,C2→C2−2C3 ⇒
[cos2θ(2+4‌sin‌4‌θ)+(1−cos2θ) (2+4‌sin‌4‌θ)]=0 ⇒[2cos2θ+4cos2θ‌sin‌4‌θ+2+4‌sin‌4‌θ −2cos2θ−4cos2θ‌sin‌4‌θ]=0 ⇒2+4‌sin‌4‌θ=0 ⇒sin‌4‌θ=−
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