Fsc Part 2 Mathematics (Complete Solutions)

# Q8 Prove the identities:

(i)     sinh2x = 2sinhxcoshx
(ii)    sech2x = 1 - tanh2x
(iii)   csch2x = coth2x - 1

Solution

(i)     sinh2x = 2sinhxcoshx
R.H.S = 2sinhxcoshx
= 2 (
 ex - e-x 2
) (
 ex + e-x 2
)
=
 2 4
(ex - e-x) (ex + e-x)
=
 1 2
(ex - e-x) (ex + e-x)
=
 1 2
(ex - e-x) (ex + e-x)
=
 (ex - e-x) (ex + e-x) 2
=
 ex + x + ex - x - e-x + x - e-x - x 2
=
 e2x + e0 - e0 - e-2x 2
=
 e2x + 1 - 1 - e-2x 2
=
 e2x - e-2x 2
= sinh2x
Hence proved R.H.S = L.H.S

(ii)    sech2x = 1 - tanh2x
R.H.S = 1 - tanh2x
= 1 - (
 ex - e-x ex + e-x
)2
= 1 -
 (ex - e-x)2 (ex + e-x)2
=
 (ex + e-x)2 - (ex - e-x)2 (ex + e-x)2
=
 e2x + e-2x + 2(ex)(e-x) - [e2x + e-2x - 2(ex)(e-x)] (ex + e-x)2
=
 e2x + e-2x + 2(ex - x) - e2x - e-2x + 2(ex - x) (ex + e-x)2
=
 2(e0) + 2(e0) (ex + e-x)2
=
 2(1) + 2(1) (ex + e-x)2
=
 2 + 2 (ex + e-x)2
=
 4 (ex + e-x)2
=
 (2)2 (ex + e-x)2
= (
 2 ex + e-x
)2
= sech2x
Hence proved R.H.S = L.H.S

(iii)   csch2x = coth2x - 1
R.H.S = coth2x - 1
= (
 ex + e-x ex - e-x
)2 - 1
=
 (ex + e-x)2 (ex - e-x)2
- 1
=
 (ex + e-x)2 - (ex - e-x)2 (ex - e-x)2
=
 e2x + e-2x + 2(ex)(e-x) - [e2x + e-2x - 2(ex)(e-x)] (ex - e-x)2
=
 e2x + e-2x + 2(ex - x) - e2x - e-2x + 2(ex - x) (ex - e-x)2
=
 2(e0) + 2(e0) (ex - e-x)2
=
 2(1) + 2(1) (ex - e-x)2
=
 2 + 2 (ex - e-x)2
=
 4 (ex - e-x)2
=
 (2)2 (ex - e-x)2
= (
 2 ex - e-x
)2
= coth2x
Hence proved R.H.S = L.H.S