Question

(1 + tan square A) + (1+ 1÷tan square A)

Answer 1

Answer:

Sec²ACosec²A

Step-by-step explanation:

(1+tan²A)+(1+1/tan²A)

{as 1+tan² A= sec²A}

=Sec²A+(tan²A+1)/tan²A

=Sec²A+Sec²A/tan²A

{as tanA = SinA/CosA=SecA/CosecA}

=Sec²A+Sec²A/Sec²A/Cosec²A

=Sec²A+Cosec²A

=1/Cos²A + 1/Sin²A

=(Sin²A+Cos²A)/(Sin²ACos²A)

{as Sin²A+Cos²A =1}

=1/Sin²ACos²A

{as 1/SinA = CosecA,

1/CosA =SecA}

=Sec²ACosec²A...

hope it helps...

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