cosA-sinA+1/cosA+sinA-1 , please solve it
Answers
Answered by
5
Hello....
✔️(1+cosA)−sinAsinA−(1−cosA)
=2cos2A2−2sinA2cosA22sinA2cosA2−2sin2A2(1+cosA)−sinAsinA−(1−cosA)
=cosA2sinA2
=2cos2A22sinA2cosA2
=1+cosAsinA
=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.....
✔️(1+cosA)−sinAsinA−(1−cosA)
=2cos2A2−2sinA2cosA22sinA2cosA2−2sin2A2(1+cosA)−sinAsinA−(1−cosA)
=cosA2sinA2
=2cos2A22sinA2cosA2
=1+cosAsinA
=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.....
feeldeadpool:
thank you so much , its my maths exam tomorrow , hope it helps
Answered by
0
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
Similar questions