Prove the following identities, where the angles involved are acute angles for which the
expressions are defined.
cosA
1+sin A
+
1 + sinA
cos A =2 sec A
Answers
Answer:
LHS =(cosec θ−cotθ)
2
=(
sinθ
1
−
sinθ
cosθ
)
2
=
sin
2
θ
(1−cosθ)
2
=
1−cos
2
θ
(1−cosθ)
2
=
1+cosθ
1−cosθ
=RHS
(ii)
LHS :
1+sinA
cosA
+
cosA
1+sinA
=
(1+sinA)(cosA)
cos
2
A+(1+sinA)
2
=
(1+sinA)(cosA)
1+sin
2
A+cos
2
A+2sinA
=
(1+sinA)(cosA)
2+2sinA
=
(1+sinA)(cosA)
2(1+sinA)
=
cosA
2
=2secA
=RHS
(iii)
LHS :
1−cosθ
tanθ
+
1−tanθ
cotθ
=
sinθ
sinθ−cosθ
cosθ
sinθ
+
cosθ
cosθ−sinθ
sinθ
cosθ
=
sinθ−cosθ
1
[
cosθ
sin
2
θ
−
sinθ
cos
2
θ
]
=[
sinθ−cosθ
1
][
sinθcosθ
sin
3
θ−cos
3
θ
]
=
sinθcosθ
sin
2
θ+cos
2
θ+sinθcosθ
=
sinθcosθ
1+sinθcosθ
=secθcosec θ+1
=RHS
(iv)
secA
1+secA
=
cosA
1
1+
cosA
1
=
1
cosA+1
=
1−cosA
(cosA+1)(1−cosA)
=
1−cosA
sin
2
A
=RHS
(v)
cosA+sinA−1
cosA−sinA−1
=
sinA
cosA
+
sinA
sinA
+
sinA
1
sinA
cosA
−
sinA
sinA
+
sinA
1
=
cosA+1−cosec A
cotA−1+cosec A
(cotA)
2
−(1−cosec A)
2
(cotA−1+cosec A)
2
=
(−1−1+2cosA)
(2cosec A−2)(cosec A+cotA)
=cosec A+cotA
=RHS
(vi)
(1−sinA)(1+sinA)
(1+sinA)(1+sinA)
=
cosA
1+sinA
=secA+tanA
=RHS
(vii)
2cos
3
θ−cosθ
sinθ−2sin
3
θ
=
cosθ(2cos
2
θ−1)
sinθ(1−2sin
2
θ)
=tanθ[
2−2sin
2
θ−1
(1−2sin
2
θ)
]
=tanθ= RHS
(viii)
(sinA+cosec A)
2
+(cosA+secA)
2
=sin
2
A+cosec
2
A+2sinA⋅cosec A+cos
2
A+sec
2
A+2sec⋅cosA
=1+(1+cot
2
A+1+tan
2
A)+2+2
=7+tan
2
A+cot
2
A
=RHS
(ix)
LHS : (cosec A−sinA)(secA−cosA)
=(
sinA
1
−sinA)(
cosA
1
−cosA)
=(
sinA
1−sin
2
A
)(
cosA
1−cos
2
A
)=
sinAcosA
(cos
2
A)(sin
2
A)
=sinAcosA
RHS :
tanA+cotA
1
=
sin
2
+cos
2
A
sinAcosA
=sinAcosA
∴LHS=RHS
(x)
LHS :
1+cot
2
A
1+tan
2
A
=
1+
sin
2
A
cos
2
A
1+
cos
2
A
sin
2
A
=
sin
2
A
1
cos
2
A
1
=
cos
2
A
sin
2
A
=tan
2
A
RHS : (
1−cotA
1−tanA
)
2
=
1+cot
2
A−2cotA
1+tan
2
−2tanA
=
cosec
2
A−2cotA
sec
2
A−2tanA
=
sin
2
A
1−2sinAcosA
cos
2
A
1−2sinAcosA
=
cos
2
A
sin
2
A
=tan
2
A
∴LHS=RHS