Question

HELP ME PEOPLE: SOLVE (1/sec^2-cos^2+1/cosec^2-sin^2 )(sin^2cos^2)

Answer 1

since 1/sec²=cos² and 1/cosec²=sin²
therefore
(1/sec²-cos²+1/cosec²-sin²)(sin²cos²)
=(cos²-cos2+sin²-sin²)(sin²cos²)
=0*sin²cos²=0

Answer 2

Answer: in comment which you did is correct ok

Step-by-step explanation:

{1/(sec²a-cos²a)+1/(cosec²a-sin²a )}(sin²a.cos²a)

= (cos²a/1-cos⁴a) + (sin²a/1-sin⁴a)(sin²a.cos²a)

= [(cos²a-sin⁴a.cos²+sin²a-sin²a.cos⁴a)/{(1-sin⁴a)(1-cos⁴a)}](sin²a.cos²a)

=[(sin²a+cos²a-sin⁴a.cos²a-sin²a.cos⁴a)/{(1-sin²a)(1+sin²a)(1-cos²a)(1+cos²a)](sin²a.cos²a)

= [{1-sin²a.cos²a(sin²a+cos²a)}/cos²a(1+sin²a)sin²a(1+cos²a)](sin²a.cos²a)

= (1-sin²a.cos²a)/(1+cos²a+sin²a+sin²a.cos²a)

=(1-sin²a.cos²a)/(2+sin²a.cos²a)