Question

(1+cotA-cosecA) (1+tanA + secA) = 2

Answer 1

hello there !!

we have
LHS= (1+cotA-cosecA)(1+tanA+secA)

=> (1+cosA/sinA-1/sinA)(1+sinA/cosA+1/cosA)

[since cotA=cosA/sinA , cosecA= 1/sinA, tanA=sinA/cosA, secA=1/cosA]

=> [ (sinA+cosA-1)/sinA ] [(cosA+sinA+1)/cosA] [taken LCM]

multiplying both the terms
we get ,

=> (sinAcosA+sin²A+sinA+cos²A+sinAcosA+cosA-cosA-sinA-1)/sinAcosA

=>( sinAcosA+sin²A+cos²A+sinAcosA-1)/sinAcosA

=> (2sinAcosA+1-1)/sinAcosA [since sin²A+cos²A=1]

=> 2sinAcosA/sinAcosA= 2
LHS =RHS

Answer 2

Hey Friend ✋✋✋

Plz forgive me for giving late answer.....

check the attachment for answer....