46 NUMBER AND ALGEBRA

Left hand side (L.H.S.)

= 1 +2sh^2 x= 1 + 2

(

ex−e−x

2

) 2

= 1 + 2

(

e^2 x−2exe−x+e−^2 x

4

)

= 1 +

e^2 x− 2 +e−^2 x

2

= 1 +

(

e^2 x+e−^2 x

2

)

−

2

2

=

e^2 x+e−^2 x

2

=ch 2x=R.H.S.

Problem 9. Show that th^2 x+sech^2 x=1.

L.H.S.=th^2 x+sech^2 x=

sh^2 x

ch^2 x

+

1

ch^2 x

=

sh^2 x+ 1

ch^2 x

Since ch^2 x−sh^2 x=1 then 1+sh^2 x=ch^2 x

Thus

sh^2 x+ 1

ch^2 x

=

ch^2 x

ch^2 x

= 1 =R.H.S.

Problem 10. GivenAex+Be−x≡4chx−5shx,

determine the values ofAandB.

Aex+Be−x≡4chx−5shx

= 4

(

ex+e−x

2

)

− 5

(

ex−e−x

2

)

=2ex+2e−x−

5

2

ex+

5

2

e−x

=−

1

2

ex+

9

2

e−x

Equating coefficients gives:A=−^12 andB= (^412)
Problem 11. If 4ex−3e−x≡Pshx+Qchx,
determine the values ofPandQ.
4ex−3e−x≡Pshx+Qchx
=P
(
ex−e−x
2
)
+Q
(
ex+e−x
2
)

P
2
ex−
P
2
e−x+
Q
2
ex+
Q
2
e−x

(
P+Q
2
)
ex+
(
Q−P
2
)
e−x
Equating coefficients gives:
4 =
P+Q
2
and− 3 =
Q−P
2
i.e. P+Q= 8 (1)
−P+Q=− 6 (2)
Adding equations (1) and (2) gives: 2Q=2, i.e.
Q= 1
Substituting in equation (1) gives:P= 7.
Now try the following exercise.
Exercise 25 Further problems on hyper-
bolic identities
In Problems 1 to 4, prove the given identities.

1. (a) ch (P−Q)≡chPchQ−shPshQ
   (b) ch 2x≡ch^2 x+sh^2 x


2. (a) cothx≡2 cosech 2x+thx
   (b) ch 2θ− 1 ≡2sh^2 θ


3. (a) th (A−B)≡

thA−thB

1 −thAthB

(b) sh 2A≡2shAchA

4. (a) sh (A+B)≡shAchB+chAshB

(b)

sh^2 x+ch^2 x− 1

2ch^2 xcoth^2 x

≡tanh^4 x

5. GivenPex−Qe−x≡6chx−2shx, findP
   andQ [P=2,Q=−4]


6. If 5ex−4e−x≡Ashx+Bchx, findAandB.
   [A=9,B=1]