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Question : 32 of 160
Marks:
+1,
-0
Solution:
Given,
|a|=|b|=|a+2b|=1 |a+2b|2=|a|2+4|b2|+4|a||b|cos‌θ
1=1+4+4‌cos‌θ cos‌θ=−1 ⇒θ=π ∴
sin‌θ+cos3θ+tan3θ =sin‌π+cos3π+tan3π =sin‌π+cos3π+tan3π =0+(−1)3+0=−1
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