Question

Prove than (1+tanA)²+(1+cotA)² =(secA+cosecA)²​

Answer 1

Let A =45

Then LHS

(1+1)^2+(1+1)^2= 2^2 +2^2= 4 + 4=8

RHS

(Root 2 + Root 2)^2= (2Root2)^2=8

LHS=RHS

OR

LHS

Open Squares

(1 + tan^2A + 2 tan A) + (1+cot^2+2cotA)

1+tan^2A = Sec^2A & 1+ cot^2A = cosec^2 A

Put & Get

Sec^2 A + Cosec^2 A + 2 (tan A + Cot A)

Tan=sin/cos Cot=cos /sin

So, tan A + Cot A= (sin^2 A + Cos ^2 A)/ sin A Cos A

1/ SinA Cos A= Cosec A Sec A

LHS= Sec^2 A + Cosec^2 A + 2 Cosec A Sec A

Open RHS AND you'll get the Same

Answer 2

Answer:

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