Question

sin^2A/cos^2A+cos^2A/sin^2A=sec^2A.cosec^2A-2

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

cosec^2A.sec^2A-2

Step-by-step explanation:

$sin^2A/cos^2A+cos^2A/sin^2A$

after applying lcm,

$sin^4+cos^4/sin^2A.cos^2A\\=(sin^2A)^2+2sin^2A.cos^2A+(cos^2A)^2/sin^2A.cos^2A$

by using the identity $a^2+b^2$$=(a+b)^2-2ab$

$(1)^2-2(sin^2A.cos^2A)/sin^2A.cos^2A\\=1/sin^2A.cos^2A-2(sin^2A.cos^2A)/sin^2A.cos^2A\\=1/sin^2A.cos^2A-2\\=1/1/sec^2A.cosec^2A-2\\=cosec^2A.sec^2A-2$