192 MATHEMATICS
Example 13 : Prove that sec A (1 – sin A)(sec A + tan A) = 1.
Solution :
LHS = sec A (1 – sin A)(sec A + tan A) =
11 sinA
(1 si n A)
cos A cos A cos A✁ ✁
☎ ✆ ✂ ☎ ✄ ✆
✝ ✞ ✝ ✞
=
2
22
(1 sin A) (1 + sin A) 1 sin A
cos A cos A✟ ✟
✠
=
2
2
cos A
1
cos A✠ = RHS
Example 14 : Prove that
cot A – cos A cosec A – 1
cot A + cos A cosec A + 1✡
Solution : LHS =
cos A cos A
cot A – cos A sin A
cot A + cos A cos A
cos A
sin A☛
☞
✌
=
cos A^1111
sin A sin A cosec A – 1
11 cosec A + 1
cos A 1 1
sin A sin A✍ ✎ ✍ ✎
✏ ☛ ✑ ✏ ☛ ✑
✒ ✓☞✒ ✓☞
✍ ✎ ✍ ✎
✏ ✌ ✑ ✏ ✌ ✑
✒ ✓ ✒ ✓
= RHS
Example 15 : Prove that
sin cos (^11) ,
sin cos 1 sec tan
✔✕ ✔✖
✗
✔✖ ✔✕ ✔✕ ✔
using the identitysec^2 ✘ = 1 + tan^2 ✘.
Solution : Since we will apply the identity involving sec ✘ and tan ✘, let us first
convert the LHS (of the identity we need to prove) in terms of sec ✘ and tan ✘ by
dividing numerator and denominator by cos ✘✙
LHS =
sin – cos + 1 tan 1 sec
sin + cos – 1 tan 1 sec✚ ✚ ✚✛ ✜ ✚
✡
✚ ✚ ✚✜ ✛ ✚
=
(tan sec ) 1 {(tan sec ) 1} (tan sec )
(tan sec ) 1 {(tan sec ) 1} (tan sec )