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Question : 47 of 120
Marks:
+1,
-0
Solution:
‌−‌ =‌| √3‌cos(20∘)−sin(20∘) |
| sin(20∘)‌cos(20∘) |
=‌| 4[‌cos(20∘)−] |
| 2‌sin(20∘)‌cos(20∘) |
[∵2‌sin‌A‌cos‌A=sin‌2‌A] =‌| 4‌sin(60∘−20∘) |
| sin‌40∘ |
[∵sin(A−B)=sin‌A‌cos‌B−cos‌A‌sin‌B]
=‌| 4‌sin‌40∘ |
| sin‌40∘ |
=4
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