I can't prove this trigonometric identity. Can somebody help me?

Hello,

let's work on the left side:

[1 /(sin²a + cos²a + tg²a)] - {sin²(90°+ a) + [cos(90°+ a) /sin a]} =

let's apply (in the first denominator) the fundamental identity sin²a + cos²a = 1:

{1 /[(sin²a + cos²a) + tg²a]} - {sin²(90°+ a) + [cos(90°+ a) /sin a]} =

[1 /(1 + tg²a)] - {sin²(90°+ a) + [cos(90°+ a) /sin a]} =

let's express the tangent in terms of sine and cosine:

{1 /[1 + (sin a /cos a)²]} - {sin²(90°+ a) + [cos(90°+ a) /sin a]} =

{1 /[1 + (sin²a /cos²a)]} - {sin²(90°+ a) + [cos(90°+ a) /sin a]} =

(letting cos²a be the common denominator)

{1 /[(cos²a + sin²a) /cos²a]} - {sin²(90°+ a) + [cos(90°+ a) /sin a]} =

let's apply the fundamental identity cos²a + sin²a = 1:

[1 /(1 /cos²a)] - {sin²(90°+ a) + [cos(90°+ a) /sin a]} =

cos²a - {sin²(90°+ a) + [cos(90°+ a) /sin a]} =

let's consider both arguments (90°+ a) as [90° - (- a)]:

cos²a - { {sin[90° - (- a)]}² + {cos[90° - (- a)] /sin a} } =

let's apply the co-function identity sin(90° - θ) = cosθ (being of course in this case θ = - a)

cos²a - {[cos(- a)]² + {cos[90° - (- a)] /sin a} } =

similarly, let's apply the co-function identity cos(90° - θ) = sinθ (being of course in this case θ = - a)

cos²a - {[cos(- a)]² + [sin(- a) /sin a]} =

let's apply the trig identity cos(- a) = cos a:

cos²a - {(cos a)² + [sin(- a) /sin a]} =

similarly, let's apply the trig identity sin(- a) = - sin a:

cos²a - {cos²a + [(- sin a) /sin a]} =

cos²a - [cos²a - (sin a /sin a)] =

(simplifying)

cos²a - (cos²a - 1) =

cos²a - cos²a + 1 =

1

(Q.E.D.)

I hope it's helpful

sin (90 + a) = cos a

cos (90 + a) = - sin a

1 + tan²a = sec²a

thus

1 / sec²a - [ cos²a - 1 ] = cos²a - cos²a + 1 = 1