Question

tan? A + cot? A = sec? A cosec? A - 2​

Answer 1

Step-by-step explanation:

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

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

We know tan A =1/cot A

=> sec^2 A+2tan A+cosec^2 A+ 2cotA

=> sec^2 A+(2tan A++2cotA)+cosec^2 A

=> sec^2 A+(2sinA/cosA++2cosA/sinA)+cosec^2 A

=> sec^2 A+(2(sin^2A+2cos^2A)/sinAcosA)+cosec^2 A

=> sec^2 A+(2/sinAcosA)+cosec^2 A

=> sec^2 A+(2secAcosecA)+cosec^2 A

=> (secA+cosecA)^2

(1+tanA)^2+(1+cotA)^2

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

[We know that 1+tan^2(A)=sec^2(A) and 1+cot^2(A)=cosec^2(A)]

So equation becomes

Sec^2(A)+cosec^2(A)+2tanA+2cotA

Sec^2(A)+cosec^2(A)+2[sinA/cosA+cosA/sinA]

Sec^2(A)+cosec^2(A)+2[sin^2(A)+cos^2(A)] /sinACosA

=sec^2(A)+cosec^2(A)+cosecAsecA

=(secA+cosecA)^2