Question

If $\tan \left(\mathrm{\theta }\right)+\frac{1}{\tan \left(\mathrm{\theta }\right)}=3$, find the value of ${\tan }^2\left(\mathrm{\theta }\right)+\frac{1}{{\tan }^2\left(\mathrm{\theta }\right)}$.

• A. $5$
• B. $9$
• C. $4$
• D. $7$

Answer 1

The correct option is D

$7$

Given: $\tan \left(\mathrm{\theta }\right)+\frac{1}{\tan \left(\mathrm{\theta }\right)}=3$

Squaring both sides:

$\therefore$ ${\left(\tan \left(\mathrm{\theta }\right)+\frac{1}{\tan \left(\mathrm{\theta }\right)}\right)}^2={\left(3\right)}^2$

$\begin{array}{rcl}\therefore {\tan }^2\left(\mathrm{\theta }\right)+\frac{1}{{\tan }^2\left(\mathrm{\theta }\right)}+2\tan \left(\mathrm{\theta }\right)\times \frac{1}{\tan \left(\mathrm{\theta }\right)}& =& 9\left[\because {\left(a+b\right)}^2=a^2+2ab+b^2\right]\\ t{\mathrm{an}}^2\left(\mathrm{\theta }\right)+\frac{1}{{\tan }^2\left(\mathrm{\theta }\right)}+2& =& 9\\ {\tan }^2\left(\mathrm{\theta }\right)+\frac{1}{{\tan }^2\left(\mathrm{\theta }\right)}& =& 7\end{array}$

Therefore, the correct option is (D).