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Question : 22 of 160
Marks:
+1,
-0
Solution:
|sin‌α−cos‌α|=3∕4 ‌⇒‌‌(sin‌α−cos‌α)2=‌‌⇒‌‌sin‌2α+cos2α−2sin‌α‌cos‌α=‌‌⇒‌‌1−‌=2sin‌α‌cos‌α‌⇒‌‌2sin‌α‌cos‌α=‌ |sin‌α+cos‌α|‌=√sin‌2α+cos2α+2sin‌α‌cos‌α‌=√1+‌=√‌=‌ | sec2α−tan‌2‌α|=|‌−‌| ‌=|‌| (1−tan‌α)2 |
| 1−tan2α |
|=|‌|‌=‌| (cos‌α−sin‌α) |
| (cos‌α+sin‌α) |
‌=‌÷‌=‌
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