√ sec²A+cose²A=tanA+cotA
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Answer:
sec^2 A= 1/cos^2A
cosec^2A= 1/sin^2A
√1/cos^2A+1/sin^2A =√cos^2A+sin^2A/sin^2A*cos^2A= 1/sinAcosA
Now
tanA+cotA = sinA/cosA+cosA/sinA
=sin^2A+cos^2A/sinAcosA
sin^2A+cos^2A=1
Thus tanA+cotA=1/sinAcosA
LHS and RHS are same. Hence we can prove that √sec^2A+cosec^2A=tanA+cotA
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