Trigonometric Ratios & Identities

$ \begin{aligned} &\text { We have, } A+B=\frac{\pi}{4}\\ &\begin{aligned} & \Rightarrow \quad \tan (A+B)=\tan \frac{\pi}{4} \\ & \Rightarrow \quad \frac{\tan A+\tan B}{1-\tan A \tan B}=1 \\ & \Rightarrow \quad \tan A+\tan B=1-\tan A \tan B \\ & \Rightarrow \quad \tan A+\tan B+\tan A \tan B=1 \\ & \Rightarrow \quad 1+\tan A+\tan B+\tan A \tan B=2 \\ & \Rightarrow \quad(1+\tan A)+\tan B(1+\tan A)=2 \\ & \Rightarrow \quad(1+\tan A)(1+\tan B)=2 \end{aligned} \end{aligned} $

According to condition,

Let, height of tower $ =h $

In $ \triangle A B D, \quad \tan 45^{\circ}=\frac{A B}{B D} \Rightarrow B D=50 \mathrm{~m} $

Now, In $ \triangle A C C^{\prime} $

$ \begin{aligned} & & \tan 30^{\circ} & =\frac{A C^{\prime}}{C^{\prime} C} \\ \Rightarrow & & \frac{1}{\sqrt{3}} & =\frac{50-h}{50} \\ \Rightarrow & & 50 & =50 \sqrt{3}-h \sqrt{3} \\ \Rightarrow & & h \sqrt{3} & =50(\sqrt{3}-1) \\ \Rightarrow & & h & =50\left(1-\frac{1}{\sqrt{3}}\right) \\ & & & =50\left(1-\frac{\sqrt{3}}{3}\right) \mathrm{m} \end{aligned} $

Here, from the figure, we see that $\theta_2=\tan ^{-1} \frac{3}{5}$

$\Rightarrow \quad \tan \theta_2=\frac{3}{5} \quad \text{.... (i)}$

$\text { In } \triangle A O C, \tan \theta_1=\frac{A C}{A O} \Rightarrow \tan \theta_1=\frac{\frac{1}{4} h}{40}=\frac{h}{160} \quad \text{.... (ii)}$

And in $\triangle A O B, \tan \left(\theta_1+\theta_2\right)=\frac{A B}{A O}=\frac{h}{40}$

$\begin{array}{lc} \Rightarrow & \frac{\tan \theta_1+\tan \theta_2}{1-\tan \theta_1 \tan \theta_2}=\frac{h}{40} \\ \Rightarrow & \frac{\frac{h}{160}+\frac{3}{5}}{1-\frac{h}{160} \times \frac{3}{5}}=\frac{h}{40} \\ \Rightarrow & h^2-200 h+6400=0 \\ \Rightarrow & (h-160)(h-40)=0 \\ \Rightarrow & h=160 \text { or } h=40 \end{array}$

Hence, height of the vertical pole is $40 \mathrm{~m}$.

Let $A E$ be the tower $T_2$ of height $80 \mathrm{~m}$ and $B C$ be the tower $T_1$ of height $60 \mathrm{~m}$. Let $A B=C D=h \mathrm{~m}$

In $\triangle E D C$,

$\begin{aligned} & \frac{E D}{C D}=\tan x \Rightarrow \frac{20}{h}=\tan x \\ & \Rightarrow \quad h=\frac{20}{\tan x} \quad \text{.... (i)} \end{aligned}$

$\begin{array}{ll} \text { In } \triangle A B C, \frac{B C}{A B}=\tan 2 x \\ \Rightarrow \frac{60}{h}=\tan 2 x \\ \Rightarrow \frac{60}{h}=\frac{2 \tan x}{1-\tan ^2 x} \\ \Rightarrow \frac{30}{h}=\frac{\tan x}{1-\tan ^2 x} \\ \Rightarrow h=\frac{30\left(1-\tan ^2 x\right)}{\tan x} \quad \text{... (ii)} \end{array}$

From Eqs. (i) and (ii), we get

$\begin{aligned} & \frac{30\left(1-\tan ^2 x\right)}{\tan x}=\frac{20}{\tan x} \\ & \Rightarrow \quad 3-3 \tan ^2 x=2 \\ & \Rightarrow \quad \tan ^2 x=\frac{1}{3} \Rightarrow \tan x= \pm \frac{1}{\sqrt{3}} \\ & \Rightarrow \quad x=\frac{\pi}{6} \text { or } \frac{5 \pi}{6} \\ \end{aligned}$

So, on putting the value of $x$ in Eq. (i), we get

$h=\frac{20}{\tan \left(\frac{\pi}{6}\right)} \text { or } h=20 \sqrt{3} \mathrm{~m}$

Given that $A, B, C \in [0, \pi]$ and $A, B, C$ are in Arithmetic Progression (AP), we know the following:

If three angles are in AP, the middle angle is the arithmetic mean of the other two. Therefore, we can represent:

$B = \frac{A + C}{2}$

We need to find the value of the expression $\frac{\sin A + \sin C}{\cos A + \cos C}$.

Using the sum-to-product identities, we have:

$\sin A + \sin C = 2 \sin\left(\frac{A + C}{2}\right) \cos\left(\frac{A - C}{2}\right)$

And

$\cos A + \cos C = 2 \cos\left(\frac{A + C}{2}\right) \cos\left(\frac{A - C}{2}\right)$

Substituting these identities into the given expression, we get:

$\frac{\sin A + \sin C}{\cos A + \cos C} = \frac{2 \sin\left(\frac{A + C}{2}\right) \cos\left(\frac{A - C}{2}\right)}{2 \cos\left(\frac{A + C}{2}\right) \cos\left(\frac{A - C}{2}\right)}$

Canceling the common terms, the expression simplifies to:

$\frac{\sin\left(\frac{A + C}{2}\right)}{\cos\left(\frac{A + C}{2}\right)}$

Since $B = \frac{A + C}{2}$, we can substitute $B$ in place of $\frac{A + C}{2}$:

$\frac{\sin B}{\cos B}$

Which is simply:

$\tan B$

Therefore, the correct answer is:

Option D: $\tan B$

${{\sin \alpha - \sin \gamma } \over {\cos \gamma - \cos \alpha }} = {{2\cos {{\alpha + \gamma } \over 2}\sin {{\alpha - \gamma } \over 2}} \over {2\sin {{\alpha + \gamma } \over 2}\,.\,\sin {{\alpha - \gamma } \over 2}}} = \cot {{\alpha + \gamma } \over 2}$

[$\because$ $\sin A - \sin B = 2\cos \left( {{{A + B} \over 2}} \right)\sin \left( {{{A - B} \over 2}} \right)$ and $\cos A - \cos B = 2\sin \left( {{{A + B} \over 2}} \right)\sin \left( {{{B - A} \over 2}} \right)$]

But $\alpha,\beta,\gamma$ are in AP

$ \Rightarrow {{\alpha + \gamma } \over 2} = \beta $

So, required value is cot $\beta$.