If A is an acute angle, what is the value of cscA, if cotA=
p
2
−1
2p
?
- Ap
2
+1
p
2
−1
- Bp
2
−1
p
2
- Cp
2
−1
p
2
+1
- Dp
2
+1
2p
Solution & Step-by-step Explanation
We know the trigonometric identity:
csc
2
A=1+cot
2
A
Given:
cotA=
p
2
−1
2p
Substituting this value into the identity:
csc
2
A=1+(
p
2
−1
2p
)
2
csc
2
A=1+
(p
2
−1)
2
4p
2
csc
2
A=
(p
2
−1)
2
(p
2
−1)
2
+4p
2
Expanding the numerator:
(p
2
−1)
2
+4p
2
=p
4
−2p
2
+1+4p
2
=p
4
+2p
2
+1=(p
2
+1)
2
Thus,
csc
2
A=
(p
2
−1)
2
(p
2
+1)
2
Taking the square root on both sides (since A is an acute angle, cscA is positive):
cscA=
p
2
−1
p
2
+1
csc
2
A=1+cot
2
A
Given:
cotA=
p
2
−1
2p
Substituting this value into the identity:
csc
2
A=1+(
p
2
−1
2p
)
2
csc
2
A=1+
(p
2
−1)
2
4p
2
csc
2
A=
(p
2
−1)
2
(p
2
−1)
2
+4p
2
Expanding the numerator:
(p
2
−1)
2
+4p
2
=p
4
−2p
2
+1+4p
2
=p
4
+2p
2
+1=(p
2
+1)
2
Thus,
csc
2
A=
(p
2
−1)
2
(p
2
+1)
2
Taking the square root on both sides (since A is an acute angle, cscA is positive):
cscA=
p
2
−1
p
2
+1