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Question : 151 of 160
Marks:
+1,
-0
Solution:
We have,
‌|| cos(A+B) | −sin‌(A+B) | cos‌2‌B |
| sin‌A | cos‌A | sin‌B |
| −cos‌A | sin‌A | cos‌B |
|=0‌⇒‌‌cos(A+B)[cos‌A‌cos‌B−sin‌Asin‌B] ‌+sin‌(A+B)[sin‌A‌cos‌B+cos‌A‌s‌i‌n‌B]‌+cos‌2‌B[sin‌2A+cos2A]=0‌⇒‌‌cos(A+B)‌cos(A+B)‌+sin‌(A+B)sin‌(A+B)+cos‌2‌B=0‌⇒cos2(A+B)+sin‌2(A+B)+cos‌2‌B=0‌⇒‌‌1+cos‌2‌B=0‌⇒‌‌1+2cos2B−1=0‌⇒‌‌cos2B=0⇒cos2B=cos2‌‌⇒‌‌B=nπ±‌=‌(2n±1)‌∴‌‌B=‌(2n+1)‌
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