Math, asked by haritharokkam1, 7 months ago

answer it pls pls question​

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Answered by subhamrout2019
1

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

Answered by kumarigademyy
0

Answer:

sorry haritha I can't......................

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