Discussion
The value of \(\frac{1-\tan^2 45°}{1+\tan^2 45°}\) is:
Answer: D
We know that \(\tan 45° = 1\).
Substitute this value into the expression:
\(\frac{1 - (1)^2}{1 + (1)^2} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0\).
Alternatively, this is the formula for \(\cos(2\theta)\), so it equals \(\cos(2 \times 45°) = \cos 90° = 0\).
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development