# What is (1+(sinx)^2)/(2+(cotx)^2)+(1+(cosx)^2)/(2+(tanx)^2) if x<`pi` /2, x>0?

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### 1 Answer

You should use the following trigonometric identities, such that:

`1 + tan^2 x = 1/(cos^2 x) => 2 + tan^2 x = 1 + 1/(cos^2 x)`

`1 + cot^2 x = 1/(sin^2 x) => 2 + cot^2 x = 1 + 1/(sin^2 x)`

Replacing ` 1 + 1/(cos^2 x)` for `2 + tan^2 x` and `1 + 1/(sin^2 x)` for `2 + cot^2 x` yields:

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

Reducing duplicate factors yields:

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

Using Pythagorean identity yields:

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

**Hence, evaluating the given expression, under the given conditions, yields **`(1 + sin^2 x)/(2 + cot^2 x) + (1 + cos^2 x)/(2 + tan^2 x) = 1.`

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