Question

If $\sin \left(\theta \right)=\frac{24}{25}$ and $\left(\theta \right)$ lies in the second quadrant , then $\sec \left(\theta \right)+\tan \left(\theta \right)=$

• A. $-3$
• B. $-5$
• C. $-7$
• D. $-9$

Answer 1

The correct option is C

$-7$

Explanation for the correct option

Step 1: Given expression

$\sin \left(\theta \right)=\frac{24}{25}$ and $\left(\theta \right)$ lies in the second quadrant

Step 2: Simplification of the given expression

In the second quadrant $\sec \left(\theta \right)$ and $\tan \left(\theta \right)$ are negative

So by given

$\Rightarrow \sin \left(\theta \right)=\frac{24}{25}$

So by trigonometry identity,

$\Rightarrow \sin \left(\theta \right)=\frac{Perpendicular}{Hypotenuse}$

Apply Pythagoras theorem

$\begin{array}{rcl}& \Rightarrow & {\left(Hypotenuse\right)}^2={\left(Perpendicular\right)}^2+{\left(Base\right)}^2\\ & \Rightarrow & {\left(25\right)}^2={\left(24\right)}^2+{\left(Base\right)}^2\\ & \Rightarrow & 625=576+{\left(Base\right)}^2\\ & \Rightarrow & 625-576={\left(Base\right)}^2\\ & \Rightarrow & 49={\left(Base\right)}^2\end{array}$

Take square root

$$\begin{gathered} \Rightarrow Base=\sqrt{49} \\ \Rightarrow Base=7 \end{gathered}$$

Step 3: Calculate the values of $\sec \left(\theta \right)+\tan \left(\theta \right)$

$\begin{array}{rcl}\therefore \sec \left(\theta \right)& =& \frac{Hypotenuse}{Base}\\ & =& -\frac{25}{7}\left[\because \sec \left(\theta \right)liesinsecondquadrant\right]\end{array}$

And

$\begin{array}{rcl}& \Rightarrow & \tan \left(\theta \right)=\frac{Perpendicular}{Base}\\ & =& -\frac{24}{7}\left[\because \tan \left(\theta \right)liesinsecondquadrant\right]\end{array}$

Put the value of $\sec \left(\theta \right)and\tan \left(\theta \right)$in

$\begin{array}{rcl}& \Rightarrow & \sec \left(\theta \right)+\tan \left(\theta \right)\\ & \Rightarrow & \frac{-25}{7}+\left(\frac{-24}{7}\right)\left(takeL.C.M\right)\\ & \Rightarrow & \frac{-25-24}{7}\\ & \Rightarrow & \frac{-49}{7}\\ & \Rightarrow & -7\end{array}$

The value of $\begin{array}{rcl}\sec \left(\theta \right)+\tan \left(\theta \right)& =& -7\end{array}$

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