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Question : 69 of 120
Marks:
+1,
-0
Solution:
sin‌x+a‌cos‌x=b
⇒(sin‌x+a‌cos‌x)2=b2
⇒sin2x+a2cos2x+2a‌sin‌x‌cos‌x=b2
⇒(1−cos2x)+a2cos2x +2a‌sin‌x‌cos‌x=b2 ......Using‌sin2‌x=1−cos2x]
⇒(a2−1)cos2x+2a‌sin‌x‌cos‌x=b2−1
⇒2a‌sin‌x‌cos‌x=b2−1+(1−a2)cos2x ........(1)
Let k=|a‌sin‌x−cos‌x|
⇒k2=(a‌sin‌x−cos‌x)2
⇒k2=a2sin2x+cos2x−2a‌sin‌x‌cos‌x
⇒k2=a2(1−cos2x)+cos2x −2a‌sin‌x‌cos‌x ......Using‌sin2‌x=1−cos2x]
⇒k2=a2+(1−a2)cos2x−2a‌sin‌x‌cos‌x
⇒k2=a2+(1−a2)cos2x−[b2−1 +(1−a2)cos2x] ... [Using equation (1)]
⇒k2=a2−b2+1
⇒k=√a2−b2+1
∴|a‌sin‌x−cos‌x|=√a2−b2+1
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