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Question : 60 of 160
Marks:
+1,
-0
Solution:
‌‌| x‌tan‌2‌x−2x‌tan‌x |
| (1−cos‌2‌x)2 |
‌‌‌=‌| x⋅−2x‌tan‌x |
| (1−(1−2sin‌2x))2 |
‌‌‌=‌| 2x‌tan‌x(−1) |
| (2sin‌2x)2 |
‌‌‌=‌‌| 2x‌tan‌x(1−1+tan2x) | | 1−tan2x |
|
| 4sin‌4x |
=‌| 2x‌tan‌x⋅tan2x |
| 4sin‌4x(1−tan2x) |
‌‌‌=‌| xtan3x |
| 2sin‌4x(1−tan2x) |
=‌| xsin‌3x |
| 2(1−tan2x)⋅cos2x⋅sin‌2x |
‌‌‌=‌⋅‌‌=‌⋅1‌‌[∵‌=1]‌=‌=‌
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