Question

Prove that

[1/(sec²x-cos²x) + 1/(cosec²x-sin²x)] sin²x cos²x =

(1-sin²x cos²x)/(2+sin²x cos²x)

Answer 1

[1/(sec²x - cos²x) + 1/(cosec²x - sin²x)] sin²x cos²x

Mulitply top and bottom of the first fraction by cos²x, giving

cos²x / (1 - cos⁴x)
= cos²x /((1-cos²x)(1+cos²x))
= cos²x / (sin²x(1+cos²x))

A similar process applied to the second fraction (mulitply by sin²x, factorise the denominator, replace 1-sin²x by cos²x) gives

sin²x / (cos²x(1+sin²x))

Now the sum of these two has to be multiplied by sin²x cos²x:
This gives
cos⁴x / (1+cos²x) + sin⁴x / (1 + sin²x)

= [cos⁴x (1+ sin²x) + sin⁴x (1 + cos²x)] / [(1+cos²x)(1+sin²x)]

The numerator is
cos⁴x + cos⁴x sin²x + sin⁴x + sin⁴x cos²x

= cos⁴x + sin⁴x + cos²x sin²x (cos²x + sin²x)

Now using the fact that cos²x + sin²x = 1,
and squaring this relationship to get
cos⁴x + sin⁴x + 2 cos²x sin²x = 1
so that
cos⁴x + sin⁴x = 1 - 2 cos²x sin²x
we find the numerator is
1 - 2 cos²x sin²x + cos²x sin²x
= 1 - cos²x sin²x

The denominator is
(1+cos²x)(1+sin²x)
= 1 + cos²x + sin²x + cos²x sin²x
= 2 + cos²x sin²x

Checking here: If x = π/6
the left side is
(1/(4/3 - 3/4 ) + 1/(4 - 1/4) *(1/4)*(3/4)
= (12/7 + 4/15)* 3/16
= 9/28 + 1/20
= 13/35

(1 - 2 sin²x cos²x) / (sinx cosx)
= (1 - 2*(1/4)*(3/4))/(1/2)*(√3) which can't equal the left side because it has √3 in it, which the left side does not. So the question as given, even with the parentheses I've added, is not correct.

However my result
(1 - cos²x sin²x)/ (2 + cos²x sin²x)
= (1 - 3/16) / (2 + 3/16)
= 13/35

So the correct result is

[1/(sec²x - cos²x) + 1/(cosec²x - sin²x)] sin²x cos²x = (1 - cos²x sin²x)/ (2 + cos²x sin²x)