# What is the simplified form of: (sinx + cosx)^2/(1+ 2*sin*x*cos x)

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The expression to be simplified is (sinx + cosx)^2/(1+ 2*sin*x*cos x)

(sinx + cosx)^2/(1+ 2*sin*x*cos x)

Use (a + b)^2 = a^2 + b^2 + 2*a*b

=> ((sin x)^2 + (cos x)^2 + 2*sin x*cos x )/(1+ 2*sin*x*cos x)

use the relation that (sin x)^2 + (cos x)^2 = 1

=> (1 + 2*sin x*cos x ))/(1+ 2*sin*x*cos x)

=> 1

**The expression (sin x + cos x)^2/(1+ 2*sin*x*cos x) simplifies to 1.**

We'll recognize to denominator the double angle identity:

2 sin x*cos x = sin (2x)

We'll expand the binomial from numerator:

(sin x)^2 + 2sin x*cos x + (cos x)^2

We'll use the Pythagorean identity and we'll get:

(sin x)^2 + (cos x)^2 = 1

The numerator will become:

1 + sin (2x)

Now, we'll re-write the fraction:

(sinx + cosx)^2/(1+ 2*sin*x*cos x) = [1 + sin (2x)]/[1 + sin (2x)]

(sinx + cosx)^2/(1+ 2*sin*x*cos x) = 1

**The simplified fraction gives:(sinx + cosx)^2/(1+ 2*sin*x*cos x) = 1**