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Question
Show that f(x) = tan–1(sinx + cosx) is an increasing function in `(0, pi/4)`
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Solution
Given that: f(x) = tan–1(sinx + cosx) in `(0, pi/4)`
Differentiating both sides w.r.t. x, we get
f'(x) = `1/(1 + (sin x + cos x)^2) * "d"/"dx" (sinx + cos x)`
⇒ f'(x) = `(1 xx (cos x - sinx))/(1 + (sinx + cosx)^2`
⇒ f'(x) = `(cosx - sinx)/(1 + sin^2x + cos^2x + 2 sin x cos x)`
⇒ f'(x) = `(cosx - sinx)/(1 + 1 + 2 sinx cosx)`
⇒ f'(x) = `(cosx - sinx)/(2 + 2 sinx cosx)`
For an increasing function f '(x) ≥ 0
∴ `(cosx - sinx)/(2 + 2 sinx cosx) ≥ 0`
⇒ cos x – sin x ≥ 0 ....`[because (2 + sin2x) ≥ "in" (0, pi/4)]`
⇒ cos x ≥ sin x, which is true for `(0, pi/4)`
Hence, the given function f(x) is an increasing function in `(0, pi/4)`.
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