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Question : 5 of 75
Marks:
+1,
-0
Solution:
‌+(tan‌x)y‌=‌IF=e∫tan‌x‌dx‌=eln‌ ‌s‌e‌c‌x‌= secx∴‌‌ solution will be
‌y secx=∫‌ secx‌dx‌=∫‌| (2+)(1+tan2) |
| (1+‌)(1−tan2‌) |
‌dx‌‌ Let ‌tan‌=t⇒‌ sec2‌‌dx=dt‌=2‌∫‌‌dt‌=‌=‌+c‌⇒y secx=‌+c‌y(‌)=‌‌⇒2y=‌+c=‌+c=‌+c‌‌=‌+c⇒c=0‌∴y=‌| 2‌tan‌ |
| secx[tan2‌+3] |
‌‌ Now, ‌f(‌)=‌| 2‌tan‌ |
| sec‌[tan2‌+3] |
=‌
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