7. If 2sin?B - cos?ß = 2, then ß is explanation
Answers
Answered by
0
Answer:
Given, sinA+cosB=a
and sinB+cosA=b
On squaring and adding, we get
sin
2
A+cos
2
B+2sinAcosB+sin
2
B+cos
2
A+2sinBcosA
=a
2
+b
2
⇒(sin
2
A+cos
2
A)+(cos
2
B+sin
2
B)+2(sinAcosB+cosAsinB)
=a
2
+b
2
⇒1+1+2sin(A+B)=a
2
+b
2
⇒sin(A+B)=
2
a
2
+b
2
−2
.
Answered by
0
Answer:
see down
Step-by-step explanation:
Given, sinA+cosB=a
and sinB+cosA=b
On squaring and adding, we get
sin
2
A+cos
2
B+2sinAcosB+sin
2
B+cos
2
A+2sinBcosA
=a
2
+b
2
⇒(sin
2
A+cos
2
A)+(cos
2
B+sin
2
B)+2(sinAcosB+cosAsinB)
=a
2
+b
2
⇒1+1+2sin(A+B)=a
2
+b
2
⇒sin(A+B)=
2
a
2
+b
2
−2
.
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