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Question : 137 of 150
Marks:
+1,
-0
Solution:
B+C=−A⇒tan(B+C)=cot‌A ⇒
| tan‌B+tan‌C |
| 1−tan‌B‌tan‌C |
=cot‌A ⇒
tan‌A‌tan‌B+tan‌B‌tan‌C+tan‌C‌tan‌A=1 ⇒
tan‌A‌tan‌B‌tan‌C(cot‌C+cot‌A+cot‌B)=1 ⇒
cot‌A+cot‌B+cot‌C=cot‌A‌cot‌B‌cot‌C Again
cos‌2‌A+cos‌2‌B+cos‌2‌C =2‌cos(A+B)‌cos(A−B)+cos‌2‌C =2‌cos(−C)‌cos(A−B)+1−2sin2C =2‌sin‌C[cos(A−B)−sin‌C]+1 =2‌sin‌C[cos(A−B)−cos(A+B)]+1 =4‌sin‌A‌sin‌B‌sin‌C+1
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