If secA−tanA=x, then the value of x is
- A1/(sec
2
A−tan
2
A) - B1/(sec
2
A+tan
2
A) - C1/[
sec
2
A−tan
2
A
] - D1/(secA+tanA)
Solution & Step-by-step Explanation
We know the fundamental trigonometric identity:
sec
2
A−tan
2
A=1
This can be factored as a difference of squares:
(secA−tanA)(secA+tanA)=1
Given that secA−tanA=x, substituting this yields:
x(secA+tanA)=1
x=
secA+tanA
1
sec
2
A−tan
2
A=1
This can be factored as a difference of squares:
(secA−tanA)(secA+tanA)=1
Given that secA−tanA=x, substituting this yields:
x(secA+tanA)=1
x=
secA+tanA
1