If [
1+sinA
cosA
]
2
=x, then the value of x is:
- Acosec A−1
cosec A+1
- Bcosec A+1
cosec A−1
- C\left[\frac{\text{cosec } A - 1}{\text{cosec } A + 1}\right]
- D\left[\frac{\text{cosec } A + 1}{\text{cosec } A - 1}\right]
Solution & Step-by-step Explanation
Given:
x=[
1+sinA
cosA
]
2
We multiply the numerator and the denominator inside the bracket by (1−sinA):
(1+sinA)(1−sinA)
cosA(1−sinA)
=
1−sin
2
A
cosA(1−sinA)
=
cos
2
A
cosA(1−sinA)
=
cosA
1−sinA
So,
x=[
cosA
1−sinA
]
2
=
cos
2
A
(1−sinA)
2
=
1−sin
2
A
(1−sinA)
2
=
(1−sinA)(1+sinA)
(1−sinA)
2
=
1+sinA
1−sinA
Now, let's convert this in terms of cosec A by dividing both numerator and denominator by sinA:
x=
sinA
1
+1
sinA
1
−1
=
cosec A+1
cosec A−1
x=[
1+sinA
cosA
]
2
We multiply the numerator and the denominator inside the bracket by (1−sinA):
(1+sinA)(1−sinA)
cosA(1−sinA)
=
1−sin
2
A
cosA(1−sinA)
=
cos
2
A
cosA(1−sinA)
=
cosA
1−sinA
So,
x=[
cosA
1−sinA
]
2
=
cos
2
A
(1−sinA)
2
=
1−sin
2
A
(1−sinA)
2
=
(1−sinA)(1+sinA)
(1−sinA)
2
=
1+sinA
1−sinA
Now, let's convert this in terms of cosec A by dividing both numerator and denominator by sinA:
x=
sinA
1
+1
sinA
1
−1
=
cosec A+1
cosec A−1