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Question : 107 of 160
Marks:
+1,
-0
Solution:
‌‌ Given, ‌sin‌A+sin‌B=√3(cos‌B−cos‌A)‌⇒‌‌sin‌A+√3‌cos‌A=√3‌cos‌B−sin‌B‌⇒‌sin‌A+‌‌cos‌A=‌‌cos‌B−‌sin‌B‌⇒cos‌sin‌A+sin‌‌‌cos‌A‌=sin‌‌‌cos‌B−cos‌sin‌B‌⇒‌‌sin‌(A+‌)=sin‌(‌−B)‌⇒‌‌A+‌=‌−B‌⇒‌‌A=−B‌‌ Now, ‌sin‌3(A)+sin‌3B‌=sin‌(−3B)+sin‌3B‌=−sin‌3B+sin‌3B‌=0‌
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