The angle of deviation of a cloud from a point h meter above a lake is \[\theta \]. The angle of depression of its reflection in the lake is \[{{45}^{\circ }}\]. The height of the cloud
(a)\[h\tan \left( {{45}^{\circ }}+\theta \right)\]
(b)\[h\cot \left( {{45}^{\circ }}-\theta \right)\]
(c)\[h\tan \left( {{45}^{\circ }}-\theta \right)\]
(d)\[h\cot \left( {{45}^{\circ }}-\theta \right)\]

Hint: At first take the angle of depression as \[\phi \], then draw the diagram according to the question given. After that apply the trigonometric ratio tan to both the triangles containing \[\theta \] and \[\phi \] and then compare them. After comparing, put the value of \[\phi \] as \[{{45}^{\circ }}\] to find what is asked.

Complete step-by-step answer:
At first we will do the question in the generalized form which is let’s suppose the angle of depression be \[\phi \]. Let’s suppose height of cloud from ground be H. So, we will draw the diagram as,

Let A be the point from where observations were taken which is h length from the surface of the lake.
Now let’s consider triangle ABC, here we will apply trigonometric ratio which is
\[\tan \theta =\]Opposite side of angle \[\theta \] / adjacent side of angle \[\theta \]
Opposite side length which is CB will be calculated as (H - h) and adjacent side will be \[\tan \theta \] is equal to \[\dfrac{H-h}{AB}\].
So, we can write AB as, \[\dfrac{H-h}{\tan \theta }\].
Now let’s consider triangle ABR, here we will apply trigonometric ratio which is,
\[\tan \phi =\] Opposite side of angle \[\phi \] / adjacent side of angle \[\phi \]
Opposite side length which is BR will be calculated as (H + h) and the adjacent side will be represented as AB.
So, the value of \[\tan \phi \] is equal to \[\dfrac{H+h}{AB}\].
So, we can write AB as, \[\dfrac{H+h}{\tan \phi }\].
Now we know that the value of AB is \[\dfrac{H-h}{\tan \theta }\] and also the value of AB is \[\dfrac{H+h}{\tan \phi }\].
So, we can equate and write it as,
\[\dfrac{H-h}{\tan \theta }=\dfrac{H+h}{\tan \phi }\]
Or, \[\dfrac{H-h}{H-h}=\dfrac{\tan \theta }{\tan \phi }\]
Now, we will apply componendo and dividendo which means that if,
\[\dfrac{a+b}{a-b}=\dfrac{c}{d}\]
Then we can write it as,
\[\dfrac{a+b+a-b}{a+b-a+b}=\dfrac{c+d}{d-d}\]
Or, \[\dfrac{2a}{2b}=\dfrac{c+d}{c-d}\]
Or, \[\dfrac{a}{b}=\dfrac{c+d}{c-d}\]
So, on applying componendo and dividendo we get,
\[\dfrac{H+h+H-h}{H+h-H+h}=\dfrac{\tan \phi +\tan \theta }{\tan \phi -\tan \theta }\]
Or, \[\dfrac{2H}{2h}=\dfrac{\tan \phi +\tan \theta }{\tan \phi -\tan \theta }\]
Or, \[\dfrac{H}{h}=\dfrac{\tan \phi +\tan \theta }{\tan \phi -\tan \theta }\]
So, \[H=h\left( \dfrac{\tan \phi +\tan \theta }{\tan \phi -\tan \theta } \right)\]
As we took it in the generalized form we got value of H like this now as we know value of \[\tan \phi \] is \[\tan {{45}^{\circ }}\] as \[\phi ={{45}^{\circ }}\]. So, the value of H will be,
\[H=h\left( \dfrac{1+\tan \theta }{1-\tan \theta } \right)\]
We know the formula that,
\[\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}\]
So, we can put x as \[{{45}^{\circ }}\] and y as \[\theta \].
So, \[\tan \left( 45+\theta \right)=\dfrac{\tan {{45}^{\circ }}+\tan \theta }{1-\tan {{45}^{\circ }}\tan \theta }\]
Or, \[\tan \left( 45+\theta \right)=\dfrac{1+\tan \theta }{1-\tan \theta }\]
So, we can write \[\dfrac{1+\tan \theta }{1-\tan \theta }\] as \[\tan \left( 45+\theta \right)\].
Hence, H is equal to \[h\tan \left( {{45}^{\circ }}+\theta \right)\].
So, the correct option is (a).

Note: While doing calculations, one should be very careful as any small mistake can result in a wrong answer. Also, the trigonometric ratios of standard angles should be remembered as they are used very often in such questions.