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Question : 22 of 150
Marks:
+1,
-0
Solution:
| 4θ(tan‌θ−2θ‌tan‌θ) |
| (1−cos‌2‌θ) |
=| 4(θ‌tan‌θ−2θ2‌tan‌θ) |
| (1−cos‌2‌θ) |
Using L' Hospital's rule
Again using L' Hospital's rule
4(sec2θ+2θsec2θ‌tan‌θ+sec2θ −4‌tan‌θ) =| −4θsec2θ−4θsec2θ−4θ2sec2θ‌tan‌θ |
| 4‌cos‌2‌θ |
==2
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