answer it pls pls question
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Step-by-step explanation:
(1+cot A-cosec A).(1+tanA+secA)= 2
L.H.S.
=(1+cosA/sinA-1/sinA).(1+sinA/cosA+1/cosA)
=(sinA+cosA-1)×(cosA+sinA+1)/sinA.cosA)
=[(sinA+cosA)^2-(1)^2]/sinA.cosA.
=(sin^2A+cos^2A+2.sinA.cosA-1)/sinA.cosA.
=( 1+2.sinA.cosA -1)/sinA.cosA.
= 2.sinA.cosA/sinA.cosA
= 2 , proved.
L.H.S = ( 1+ cota-coseca) ( 1+ tana -seca)
= (1+cosa/sina-1/sina) (1+sina/cosa-1/cosa)
= (sina+cosa-1/sina) (cosa+sina-1/cosa)
= ((sina+cosa) - (1)/sina) ((cosa+sina) - (1)/cosa)
= ((sina+cosa)raise to power2 - (1)raise to power 2)/sinacosa
= sin2a+cos2a+2sinacosa-1/sinacosa
= 1+2sinacosa-1/sinacosa (because sin2a+cos2a=1)
= 2sinacosa/sinacosa
=2
=R.H.S
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sorry haritha I can't......................
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