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Question : 20 of 160
Marks:
+1,
-0
Solution:
‌‌| sin‌θ+sin‌3θ |
| cos‌θ+cos‌3‌θ |
‌=‌| sin‌θ+3sin‌θ−4sin‌3θ |
| cos‌θ+4cos3θ−3‌cos‌θ |
‌=‌| 4sin‌θ−4sin‌3θ |
| 4cos3θ−2‌cos‌θ |
‌=‌| 4sin‌θ⋅(1−sin‌2θ) |
| 2‌cos‌θ(2cos2θ−1) |
‌=‌| 4sin‌θ⋅cos2θ |
| 2‌cos‌θ⋅cos‌2‌θ |
‌=‌| 2sin‌θ‌cos‌θ |
| cos‌2‌θ |
‌=‌=tan‌2‌θ
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