We have to prove that (cos x)^2 * (cos y)^2 + (sin x)^2 *
(sin y) ^2 + (sin x)^2 * (cos y)^2 + (sin y)^2 * (cos x)^2 =
1
Now sin (x+y) = (sin x)*(cos y) + (cos x)*(sin y) and cos
(x+ y) = (cos x)*(cos y) - (sin x)*(sin y)
(cos x)^2 * (cos
y)^2 + (sin x)^2 * (sin y) ^2 + (sin x)^2 * (cos y)^2 + (sin y)^2 * (cos
x)^2
=> [(cos x)*(cos y)]^2 + [(sin x)*(sin y)]^2 +
[(sin x)*(cos y)]^2 + [(sin y)*(cos x)]^2
Now we use the
relation a^2 + b^2 = (a+b)^2 - 2ab
=> [cos (x+y)]^2
+ [sin (x+y)]^2 - 2*(sin x)(cos x)(sin y)(cosy) + 2*(sin x)(cos x)(sin
y)(cosy)
We know (cos x)^2 + ( sin x)^2 =
1
=> [cos (x+y)]^2 + [sin
(x+y)]^2
=>
1
Therefore (cos x)^2 * (cos y)^2 + (sin x)^2
* (sin y) ^2 + (sin x)^2 * (cos y)^2 + (sin y)^2 * (cos x)^2 =
1
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