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Question : 22 of 160
Marks:
+1,
-0
Solution:
We have,
‌+‌ ‌=‌+‌‌=‌+‌‌[∵sin‌(π+θ)=−sin‌θ‌ and ‌cos(2π−θ)=cos‌θ]
‌=‌| √3sin‌70∘−cos‌70∘ |
| sin‌70∘‌cos‌70∘ |
‌=‌| 2×2[√3∕2sin‌70∘−1∕2‌cos‌70∘] |
| 2sin‌70∘‌cos‌70∘ |
‌=‌| 4[cos‌30∘‌s‌i‌n‌70∘−sin‌30∘‌cos‌70∘] |
| sin‌140∘ |
[∵2sin‌x‌cos‌x=sin‌2x] ‌=‌| 4sin‌(70∘−30∘) |
| sin‌140∘ |
[∵sin‌(x−y)=sin‌x‌cos‌x−cos‌x‌s‌i‌n‌y]‌=‌| 4sin‌40∘ |
| sin‌(180−40) |
.‌=‌=4
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