2sec
2
A is equal to:
- A(1−tanA)
2
−(1+tanA)
2 - B(1−tanA)
2
+(1+tanA)
2
- C(1−tanA)
2
−(1+tanA)
2
- D(1−tanA)
2
+(1+tanA)
2
Solution & Step-by-step Explanation
Let's expand option D:
(1−tanA)
2
+(1+tanA)
2
Expanding each square term:
(1−2tanA+tan
2
A)+(1+2tanA+tan
2
A)
Combining like terms, the −2tanA and +2tanA cancel each other out:
=1+tan
2
A+1+tan
2
A
=2(1+tan
2
A)
From the basic trigonometric identity, 1+tan
2
A=sec
2
A:
=2sec
2
A
Hence, option D is the correct answer.
(1−tanA)
2
+(1+tanA)
2
Expanding each square term:
(1−2tanA+tan
2
A)+(1+2tanA+tan
2
A)
Combining like terms, the −2tanA and +2tanA cancel each other out:
=1+tan
2
A+1+tan
2
A
=2(1+tan
2
A)
From the basic trigonometric identity, 1+tan
2
A=sec
2
A:
=2sec
2
A
Hence, option D is the correct answer.