Angle of elevation of the top of the tower from 3
points (collinear) A, B and C on a road leading to the
foot of the tower are 30°, 45° and 60°, respectively.
The ratio of AB and BC is
Given \(A,B>0\) and \(A+B=\dfrac{\pi}{6}\). Using
\[
\tan A+\tan B=\frac{\sin(A+B)}{\cos A\cos B},
\]
with \(\sin(A+B)=\sin\frac{\pi}{6}=\dfrac12\). To minimize \(\tan A+\tan B\), maximize \(\cos A\cos B\) subject to \(A+B=\dfrac{\pi}{6}\).
The product \(\cos A\cos B\) (with fixed sum) is maximized at \(A=B=\dfrac{\pi}{12}\). Thus
\[
\cos A\cos B\le \cos^2\!\frac{\pi}{12}=\frac{1+\cos\frac{\pi}{6}}{2}
=\frac{1+\frac{\sqrt3}{2}}{2}=\frac{2+\sqrt3}{4}.
\]
Hence
\[
\min(\tan A+\tan B)=\frac{\frac12}{\frac{2+\sqrt3}{4}}
=\frac{2}{2+\sqrt3}
=\boxed{\,4-2\sqrt3\,}.
\]
Given $\,\csc\theta-\cot\theta=2\,$ and the identity $\;(\csc\theta-\cot\theta)(\csc\theta+\cot\theta)=1\;$, we get
$$\csc\theta+\cot\theta=\frac{1}{2}.$$
Adding the two equations:
$$(\csc\theta-\cot\theta)+(\csc\theta+\cot\theta) $$ $$=2+\frac{1}{2}$$ $$\;\Rightarrow\;2\,\csc\theta=\frac{5}{2}.$$
Hence
$$\csc\theta=\frac{5}{4}.$$
If $a_1, a_2, a_3,...a_n$, are in Arithmetic Progression
with common difference d, then the sum $(sind) (cosec a_1 . cosec a_2+cosec a_2.cosec a_2+...+cosec a_{n-1}.cosec a_n)$ is equal to
In a right angled triangle, the hypotenuse is four times the perpendicular drawn to it from the opposite vertex. The value of one of the acute angles is
If $\prod ^n_{i=1}\tan ({{\alpha}}_i)=1\, \forall{{\alpha}}_i\, \in\Bigg{[}0,\, \frac{\pi}{2}\Bigg{]}$ where i=1,2,3,...,n. Then maximum value of $\prod ^n_{i=1}\sin ({{\alpha}}_i)$.