CosA-sinA +1 /CosA+sinA-1 = SinA/1-cosA
Answers
Answered by
0
Answer:
Step-by-step explanation:
(cosA-sinA)+1/(cosA+sinA)-1
doing the rationalization of the denominator (cosA+sinA)-1
(cosA-sinA)+1/(cosA+sinA)-1 × [(cosA+sinA)+1/(cosA+sinA)+1]
on multiplying we get -:
=[cos2A-sin2A+2cosA+1]/2sinAcosA
=[cos2A-sin2A+2cosA+sin2A+cos2A]/2sinAcosA
=[2cos2A+2cosA]/2sinAcosA
=[cosA+cos2A]/sinAcosA
This can also be written as-:
=cosA/sinAcosA +cos2A/sinAcosA
CosecA+cotA =R.H.S
Similar questions