(1+cot A -coses A) (1+tan A+sec A) =2

1.	(1+cot A -coses A) (1+tan A+sec A) =2
Answer» (1/1+cosA/1-1/sinA )(1/1+sinA/cosA+1/cosA)[(SinA+cosA-1)÷sinA] [(cosA+sinA+1)÷cosA][(sinA+cosA)^2 - 1]÷sinAcosA[SinA^2+ cosA^2+ 2(sinAcosA)-1]÷sinAcosA1+ 2(sinAcosA) -1÷sinAcosA =2 Yaha pr +1 ,-1cncl ho jayenge or sinAcosA dono cncl ho jayenge or last me 2 bach jayega.. LHS=RHS

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