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Question : 63 of 160
Marks:
+1,
-0
Solution:
| tan‌x+4‌tan‌2‌x−3‌tan‌3‌x |
| x2‌tan‌x |
=[tan‌x+4()−(| 3‌tan‌x−tan3x | | 1−3tan2x |
) |
| x2‌tan‌x |
] =[| tan‌x{1+−} |
| x2‌tan‌x |
] =[| −16tan2x |
| x2(1−tan2x)(1−3tan2x) |
] =−16()2 =−16(1)2.=−16
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