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  4. If f(θ)=(1- sin 2 θ+ cos 2 θ/2 cos 2 θ) then value of f(11°) ⋅ f(34°) is.

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Q. If $f(\theta)=\frac{1-\sin 2 \theta+\cos 2 \theta}{2 \cos 2 \theta}$ then value of $f\left(11^{\circ}\right) \cdot f\left(34^{\circ}\right)$ is_______.

Trigonometric Functions

Solution:

$f(\theta) =\frac{1-\sin 2 \theta+\cos 2 \theta}{2 \cos 2 \theta}$
$=\frac{(\cos \theta-\sin \theta)^{2}+\left(\cos ^{2} \theta-\sin ^{2} \theta\right)}{2(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)}$
$=\frac{\cos \theta}{\cos \theta+\sin \theta}=\frac{1}{1+\tan \theta}$
$f\left(11^{\circ}\right) \cdot f\left(34^{\circ}\right)=\frac{1}{\left(1+\tan 11^{\circ}\right)} \cdot \frac{1}{\left(1+\tan 34^{\circ}\right)}$
$=\frac{1}{\left(1+\tan 11^{\circ}\right)} \cdot \frac{1}{\left(1+\tan \left(45^{\circ}-11^{\circ}\right)\right)}$
$=\frac{1}{\left(1+\tan 11^{\circ}\right)} \cdot \frac{1}{1+\frac{1-\tan 11^{\circ}}{1+\tan 11^{\circ}}}$
$=\frac{1}{\left(1+\tan 11^{\circ}\right)} \cdot \frac{1+\tan 11^{\circ}}{2}=\frac{1}{2}$