8.
Prove that cosh-¹x = log(x+√x²-1)
Answers
Answered by
1
Step-by-step explanation:
Let
y
=
cosh
−
1
x
then by definition
x
=
cosh
y
=
e
y
+
e
−
y
2
2
x
=
e
y
+
e
−
y
e
y
−
2
x
+
e
−
y
=
0
e
y
−
2
x
+
1
e
y
=
0
e
2
y
−
2
x
e
y
+
1
=
0
Let
u
=
e
y
then we have
u
2
−
2
x
u
+
1
=
0
--> Now use quadratic formula to solve
u
=
2
x
±
√
4
x
2
−
4
2
e
y
=
2
x
±
√
4
x
2
−
4
2
e
y
=
2
x
±
√
4
(
x
2
−
1
)
2
2
e
y
=
2
x
±
2
√
x
2
−
1
e
y
=
x
±
√
x
2
−
1
ln
e
y
=
ln
(
x
±
√
x
2
−
1
)
y
=
ln
(
x
+
√
x
2
−
1
)
cosh
−
1
x
=
ln
(
x
+
√
x
2
−
1
)
Answered by
2
Answer:
letx= cosh x
log e = 1
log e^(u) = u log e
put the valureof sin hx and cos hx
sinhx = 1/2 (e^x- e^-x)
coshx = 1/2 ( e^x + e-^x)
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