Question

If $\sec \left(\theta \right)=m$ and $\tan \left(\theta \right)=n$, then $\frac{1}{m}\left[\left(m+n\right)+\frac{1}{\left(m+n\right)}\right]$ is equals to

• A. $2$
• B. $2m$
• C. $2n$
• D. $mn$

Answer 1

The correct option is A

$2$

Explanation for the correct option

Substitute the value of $\sec \left(\theta \right)=m$ and $\tan \left(\theta \right)=n$ in $\frac{1}{\mathrm{m}}[(\mathrm{m}+\mathrm{n})+\frac{1}{(\mathrm{m}+\mathrm{n})}]$, then

$\begin{array}{c}\frac{1}{\mathrm{m}}[(\mathrm{m}+\mathrm{n})+\frac{1}{(\mathrm{m}+\mathrm{n})}]=\frac{1}{\sec \left(\theta \right)}\left[\sec \left(\theta \right)+\tan \left(\theta \right)+\frac{1}{\sec \left(\theta \right)+\tan \left(\theta \right)}\right]\\ =\frac{1}{\sec \left(\theta \right)}\left[\frac{{\sec }^2\left(\theta \right)+{\tan }^2\left(\theta \right)+2\tan \left(\theta \right)\sec \left(\theta \right)+1}{\sec \left(\theta \right)+\tan \left(\theta \right)}\right]\end{array}$

Put trigonometry identity $\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{2}}\left(\theta \right)\mathbf{=}\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}}\left(\theta \right)\mathbf{-}\mathbf{1}$

$\begin{array}{rcl}\frac{1}{\mathrm{m}}[(\mathrm{m}+\mathrm{n})+\frac{1}{(\mathrm{m}+\mathrm{n})}]& =& \frac{1}{\sec \left(\theta \right)}\left[\frac{{\sec }^2\left(\theta \right)+{\sec }^2\left(\theta \right)-1+2\tan \left(\theta \right)\sec \left(\theta \right)+1}{\sec \left(\theta \right)+\tan \left(\theta \right)}\right]\\ & =& \frac{2{\sec }^2\left(\theta \right)+2\sec \left(\theta \right)\tan \left(\theta \right)}{\sec \left(\theta \right)\left[\sec \left(\theta \right)+\tan \left(\theta \right)\right]}\end{array}$

Take common $2\sec \left(\theta \right)$ in the numerator,

$\begin{array}{rcl}\frac{1}{\mathrm{m}}[(\mathrm{m}+\mathrm{n})+\frac{1}{(\mathrm{m}+\mathrm{n})}]& =& \frac{2\sec \left(\theta \right)\left[\sec \left(\theta \right)+\tan \left(\theta \right)\right]}{\sec \left(\theta \right)\left[\sec \left(\theta \right)+\tan \left(\theta \right)\right]}\\ & =& 2\end{array}$

The value of $\frac{1}{m}\left[\left(m+n\right)+\frac{1}{\left(m+n\right)}\right]$ is equal to $2$

Hence, option $\left(\mathit{A}\right)$ is correct.