Question

$\tan {70}^{°},\tan {50}^{°}+\tan {20}^{°},\tan {20}^{°}$are in

• A. AP
• B. GP
• C. HP
• D. None of these

Answer 1

The correct option is A

AP

Explanation for correct option

Step 1: Given Data

$\tan {70}^{°},\tan {50}^{°}+\tan {20}^{°},\tan {20}^{°}$

We know that,

$$\begin{gathered} \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \\ \therefore \tan 70°=\tan \left(50°+20°\right)\left[\because 70°=50°+20°\right] \end{gathered}$$

Step 2: taking $\tan$ on both sides:

$$\begin{gathered} \Rightarrow \tan 70°=\frac{\tan 50°+\tan 20°}{1-\tan 50°\tan 20°} \\ \Rightarrow \tan 70°\left(1-\tan 50°\tan 20°\right)=\tan 50°+\tan 20° \\ \Rightarrow \tan 70°-\tan 70°\tan 50°\tan 20°=\tan 50°+\tan 20°-\left(1\right) \end{gathered}$$

Now we can rewrite it as:

$$\begin{gathered} \Rightarrow \tan \left(70°\right)=\tan \left(90°-20°\right) \\ \Rightarrow =\cot 20° \\ \Rightarrow =\frac{1}{\tan 20°}\left[\because \tan \left(90°-\theta \right)=\cot \theta ,\cot \theta =\frac{1}{\tan \theta }\right] \end{gathered}$$

Therefore, from equation $\left(1\right)$,

$\tan 70°-\frac{1}{\tan 20°}\times \tan 50°\times \tan 20°=\tan 50°+\tan 20°$

$\Rightarrow \tan 70°-\tan 50°=\tan 50°+\tan 20°$

Step 3: simplifying.

$\tan 70°=\tan 50°+\tan 50°+\tan 20°$

$\tan 70°=2\tan 50°+\tan 20°$

Adding $\tan 20°$on both sides,

$$\begin{gathered} \tan 70°+\tan 20°=2\tan 50°+\tan 20°+\tan 20° \\ \Rightarrow \tan 70°+\tan 20°=2\left(\tan 50°+\tan 20°\right) \end{gathered}$$

Step 4: Expanding R.H.S,

$\Rightarrow \tan 70°+\tan 20°=\left(\tan 50°+\tan 20°\right)+\left(\tan 50°+\tan 20°\right)$

Simplifying further,

$\Rightarrow \tan 70°-\left(\tan 50°+\tan 20°\right)=\left(\tan 50°+\tan 20°\right)-\tan 20°$

In A.P, we know that,

$\mathrm{a}-\mathrm{b}=\mathrm{b}-\mathrm{c}(\mathrm{when}\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P})$

$\therefore \tan 70°,\left[\tan \left(50°\right)+\tan 20°\right],\tan 20°$ are in A.P.

Hence, the correct answer is option (A).