Tricky Trigonometry - 1

$\begin{aligned} f(\theta) & = \frac {1-\sin 2\theta + \cos 2\theta}{2\cos 2 \theta} & \small \blue{\text{Let }t = \tan \theta \text{ and using half-}} \\ & = \frac {1-\frac {2t}{1+t^2}+\frac {1-t^2}{1+t^2}}{2\times \frac {1-t^2}{1+t^2}} & \small \blue{\text{angle tangent substitution.}} \\ & = \frac 1{1+t} = \frac 1{1+\tan \theta} \end{aligned}$

Then

$\begin{aligned} f(11^\circ) f(34^\circ) & = \frac 1{(1+\tan 11^\circ)(1+\tan 34^\circ)} \\ & = \frac 1{(1+\tan 11^\circ)(1+\tan (45^\circ - 11^\circ))} \\ & = \frac 1{(1+\tan 11^\circ)\left(1+\frac {1-\tan 11^\circ}{1+\tan 11^\circ}\right)} \\ & = \frac 12 = \boxed{0.5} \end{aligned}$

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Reference: Half-angle tangent substitution

The given function can be simplified to $f(\theta)=$ $\dfrac{1}{1+tan \theta}$ . So $f(\theta)\times f(\dfrac{π}{4}-\theta) =0.5$ for all values of $\theta$ . Hence the required answer is $0.5$