Maths

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) =
=
=
1 1 ✁ ✁ sin A
✂ (1 ✄
sin ☎ A)
✆ ☎ ✆
✝ ✞ ✝ ✞
cos A cos A cos A
(1 sin A)(1 + sin A) 1 sin ✟
✟ A
2
cos A 1
2 2
✠
cos A cos A
2 = RHS
cos A ✠
2
Example 14 : Prove that
cot A – cos A cosec A – 1
cot A + cos A cosec A + 1
✡
Solution : LHS =
cot A – cos A
cot A + cos A
☞
cos A
sin A
cos A
sin A
☛ cos A
✌ cos A
=
1 ✍ ✎ ✍ ✎ 1
cos ☛
☛
✏ ✑ ✏ ✑
A 1 1
sin A sin A cosec ✒ A ✓ – ✒ 1
✓
☞
☞
✍ ✎ ✍ ✎
1 1 cosec A + 1
✑
✏ ✑ ✏ ✌
✌
cos A 1 1
sin A sin A
= RHS
✒ ✓ ✒ ✓
Example 15 : Prove that sin cos 1 1 ✔ ✔
,
✖ ✕ ✗
✔ ✖ ✔ ✕ ✔ ✕ ✔ sin cos 1 sec tan
sec 2 = 1 + tan 2 ✘.
✘
using the identity
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 )
✔ ✕ ✔ ✖ ✔ ✕ ✔ ✖ ✔✕ ✔