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Question : 62 of 70
Marks:
+1,
-0
Solution:
‌{tan−1(sec‌x−tan‌x)} =‌{tan−1(‌−‌)} =‌{tan−1(‌)}=‌ {tan−1(‌| sin2+cos2−2‌sin‌‌cos‌)] |
| cos2‌−sin2‌ |
)] [∵sin2x+cos2x=1 and
sin‌x=2‌sin‌‌cos‌ and
cos‌A=cos2‌−sin2‌] =‌{tan−1(‌| (cos‌−sin‌)2 |
| (cos‌+sin‌)(cos‌−sin‌) |
)} [∵a2−b2=(a+b)(a−b) ‌‌ and
(a−b)2=a2+b2−2ab] =‌{tan−1(‌| cos‌−sin‌ |
| cos‌+sin‌) |
} =‌{tan−1(‌)} [| ∵‌ divided by ‌cos‌‌ in denominator ‌ |
| ‌ and numerator ‌ |
]. =‌{tan−1(‌| tan‌−tan‌ |
| 1+tan‌⋅tan‌ |
)} =‌{tan−1‌tan(‌−‌)}=‌(‌−‌) =−‌
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