Question

√sec²A + cosec²B =tan A + cot B​

Answer 1

ANSWER:-

LHS:-

Root over  :- sec²+cosec²

                   =1/cos² + 1/sin²

                   =(sin²+cos²)/(cos²sin²)             [   LCM]

                  =root over  1/cos^2sin^2

                  =1/cosAsinA    ....(1)

 

RHS:-

               tanA+cotA

             =sin/cos +  cos/sin

             = (sin²+cos²)/cosAsinA

             =1/cosAsinA........(2)

FROM 1 & 2.

LHS=RHS

Answer 2

Step-by-step explanation:

sec^2A=1+tan^2A(i used A instead of theta as i was unable to find symbol)

cosec^2A=1+cot^2A

so √sec^2A+Cosec^2A=√tan^2A+2+cot^2A

√sec^2A+cosec^A=√tan^2A+2TanA.cotA+cot^2A

√sec^2A+cosec^A=TanA+cotA