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Question : 20 of 160
Marks:
+1,
-0
Solution:
cos210°+cos250°−sin‌40°‌sin‌80° =[2cos210°+2cos250°−2‌sin‌40°‌sin‌80°] ∵{2cos2A=1+cos‌2‌A and
2‌sin‌A‌sin‌B =cos(A−B)−cos(A+B)} [∵cos(−θ)=cos‌θ,cos(120°)=−] =[+2‌cos‌60°.cos‌40°−cos‌40°] =[+2×‌cos‌40°−cos‌40°] =[+cos‌40°−cos‌40°]=×=
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