Question

cosA-sinA+1/cosA+sinA-1 , please solve it

Answer 1

Hello....

✔️(1+cosA)−sinAsinA−(1−cosA)


=2cos2A2−2sinA2cosA22sinA2cosA2−2sin2A2(1+cos⁡A)−sin⁡Asin⁡A−(1−cos⁡A)


=cos⁡A2sin⁡A2


=2cos2A22sin⁡A2cos⁡A2


=1+cos⁡Asin⁡A


=cosecA+cotA




➖➖➖➖Alternatively ➖➖➖➖



✔️✔️ (cosA-sinA+1)/(cosA+sinA-1)


={sinA(cosA-sinA+1)}/{sinA(cosA+sinA-1)}


➖(multiplying numerator and denominator by sinA)➖



={sinA(1+cosA)-sin^2A}/{sinA(cosA+sinA-1)}


={sinA(1+cosA)-(1-cos^2A)}/{sinA(cosA+sinA-1)}


={sinA(1+cosA)-(1+cosA)(1-cosA)}/{sinA(cosA+sinA-1)}


={(1+cosA)(sinA-1+cosA)}/{sinA(cosA+sinA-1)}


={(1+cosA)(cosA+sinA-1)}/{sinA(cosA+sinA-1)}


=(1+cosA)/sinA


=1/sinA+cosA/sinA


=cosecA+cotA



thank you.....

Answer 2

Answer:

Lets take and solve :-

(cosA-sinA)+1/(cosA+sinA)-1

taking conjugate

(cosA-sinA)+1/(cosA+sinA)-1  ×  (cosA+sinA)+1/(cosA+sinA)+1

multiplying we get

=cos2A-sin2A+2cosA+1/2sinAcosA

=cos2A-sin2A+2cosA+sin2A+cos2A/2sinAcosA                  

on solving we get

=2cos2A+2cosA/2sinAcosA

=cosA+cos2A/sinAcosA

we can alsom write in the form of

cosA/sinAcosA  +cos2A/sinAcosA

on  solving we get

CosecA+cotA =R.H.S