Math, asked by raghavkashyap1972, 8 months ago

Prove that ( 1 + CotA - CosecA) ( 1 + TanA + SecA) =2

Answers

Answered by saipavan018

Answer:

Hello dude

Here is your answer

L.H.S = ( 1+ cota-coseca) ( 1+ tana -seca)

Step-by-step explanation:

= (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

=R.H.S

Hence Proved

Hope Helpful ✌️

Answered by adityachoudhary2956

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◇ [1+cotA-CosecA]*[1+tanA+secA]

= 1 + tanA + secA + cotA + 1 + cosecA - (1/sinA) - (1/cosA)

- secAcosecA

= 1 + tanA + secA + cotA + 1 + cosecA - cosecA - secA - secAcosecA

= 2 + tanA + cotA - secAcosecA

= 2 + [(sin^2A + cos^2A)/(sinAcosA)] - secAcosecA

= 2 + 1/(sinAcosA) - secAcosecA

= 2 + secAcosecA - secAcosecA

= 2

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