If
1+sinA
1−sinA
=x, then x is:
- A(cscA−cotA)
2 - BsecA−tanA
- C(secA−tanA)
2 - DcscA−cotA
Solution & Step-by-step Explanation
Given expression:
x=
1+sinA
1−sinA
Rationalizing the denominator by multiplying the numerator and denominator by (1−sinA):
x=
(1+sinA)(1−sinA)
(1−sinA)(1−sinA)
x=
1−sin
2
A
(1−sinA)
2
Since 1−sin
2
A=cos
2
A:
x=
cos
2
A
(1−sinA)
2
x=(
cosA
1−sinA
)
2
x=(
cosA
1
−
cosA
sinA
)
2
x=(secA−tanA)
2
x=
1+sinA
1−sinA
Rationalizing the denominator by multiplying the numerator and denominator by (1−sinA):
x=
(1+sinA)(1−sinA)
(1−sinA)(1−sinA)
x=
1−sin
2
A
(1−sinA)
2
Since 1−sin
2
A=cos
2
A:
x=
cos
2
A
(1−sinA)
2
x=(
cosA
1−sinA
)
2
x=(
cosA
1
−
cosA
sinA
)
2
x=(secA−tanA)
2