if in ABC, LC = 90, then sin (A+B) =
Answers
Answer:
ANSWER
Given that, In ΔABC,∠C=90
0
A+B+C=180
0
A+B=180
0
−90
0
A+B=90
0
........(1)
Then,
a
2
−b
2
a
2
+b
2
sin(A−B)
Using sine rule, and we get,
sinA
a
=
sinB
b
=
sinC
c
=k
(ksinA)
2
−(ksinB)
2
(ksinA)
2
+(ksinB)
2
sin(A−B)
=
sin
2
A−sin
2
B
sin
2
A+sin
2
B
sin(A−B)
=
2
1−cos2A
−
2
1−cos2B
2
1−cos2A
+
2
1−cos2B
sin(A−B)∴cos2θ=1−2sin
2
θ
=
1−cos2A−1+cos2B
1−cos2A+1−cos2B
sin(A−B)
=
cos2B−cos2A
2−cos2A−cos2B
sin(A−B)
=
cos2B−cos2A
2−(cos2A+cos2B)
sin(A−B)
=
2sin
2
(2B+2A)
sin
2
(2A−2B)
2−(2cos
2
(2A+2B)
cos
2
(2A−2B)
)
sin(A−B)
=
2sin(A+B)sin(A−B)
2−(2cos(A+B)cos(A−B))
sin(A−B)
=
2sin(A+B)sin(A−B)
2−(2cos(A+B)cos(A−B))
sin(A−B)
=
2sin(A+B)
2−(2cos(A+B)cos(A−B))
Now, by equation (1) and we get,
=
2sin90
0
2−(2cos90
0
cos(A−B))
=
2
2−0
=1
Hence, this is the answer.