What is the value of [cos (90° + A) ÷ sec (270° – A)] + [sin (270° + A) ÷ cosec (630° – A)]?
Show answer & explanation
Correct answer: D
Explanation
The correct answer is option 4 i.e. 1
[cos (90° + A) ÷ sec (270° – A)] + [sin (270° + A) ÷ cosec (630° – A)]
= -sinA/-cosecA + (-cosA)/-secA
= sin2A + cos2A = 1