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Question : 40 of 50
Marks:
+1,
-0
Solution:
‌ ‌‌{tan−1(sec‌x+tan‌x)}=? ⇒
?‌‌=‌{tan−1(‌+‌)} ‌‌=‌{tan−1(‌)} ‌‌=‌ [tan−1(‌| cos2+sin2+2‌sin‌‌cos‌) |
| cos2‌−sin2‌ |
)} =‌{tan−1‌| (cos‌+sin‌)2 |
| (cos‌+sin‌)(cos‌−sin‌) |
} =‌{tan−1(‌| cos‌+sin‌ |
| cos‌−sin‌ |
)} =‌{tan−1‌| cos‌(1+tan‌) |
| cos‌(1−tan‌) |
} =‌{tan−1‌| tan‌+tan‌ |
| 1−tan‌‌tan‌ |
} [∵tan‌=1] =‌{tan−1‌tan(‌+‌)} [∵tan(A+B)=‌| tan‌A+tan‌B |
| 1−tan‌A‌tan‌B |
] =‌(‌+‌)=0+‌=‌
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