Correct option is C
Given:
1−cosA1+cosA
Concept:
sinA1 = cosec A
sinAcosA = cot A
Solution:
1−cosA1+cosA
We multiply both the numerator and the denominator by 1+ cosA
(1−cosA)(1+cosA)(1+cosA)2
The denominator is now a difference of squares:
(1−cosA) (1+cosA) = 12−cos2A=sin2A
sin2A(1+cosA)2
sinA1+cosA
sinA1+sinAcosA
cosecA + cotA
1+cosA1 + \cos AThus, the correct option is (c) cosecA + cotA