Question

If \[\sin \theta = \dfrac{{24}}{{25}}\] and $\theta $ lies in the second quadrant then $\sec \theta + \tan \theta = $
A. -3
B. -5
C. -7
D. -9

Answer 1

Hint: There are four quadrants in a graph or XY plane, In each quadrant all trigonometric ratios have different limits within that limit they are either positive or negative only in the first quadrant all trigonometric ratios are positive and in other quadrants some angles are positive and rest are negative.
In this we have $\theta $ in second quadrant so we will first find out value of $\tan \theta $ and $\sec \theta $

Formula used: Pythagoras theorem, ${H^2} = {P^2} + {B^2}$
Square of the largest side (hypotenuse) is equal to the sum of squares of the other two sides (perpendicular and base).

Complete step-by-step solution:
Firstly it is mentioned in the question $\theta $ lies in the second quadrant.
In the second quadrant only $\sin \theta $ and $\cos ec\theta $ are positive.
All the other trigonometric ratios are negative.
We know that,
\[\sin \theta = \dfrac{{24}}{{25}}\] and $\sin \theta = \dfrac{{perpendicular}}{{hypotenuse}}$
So we have ,
$\dfrac{{24}}{{25}} = \dfrac{{perpendicular}}{{hypotenuse}}$
Therefore , perpendicular = 24 and hypotenuse = 25.
So, let’s find out base using Pythagoras theorem,
${H^2} = {P^2} + {B^2}$
Putting the value of hypotenuse and perpendicular we will solve for base and get its value.
Therefore,
$\Rightarrow$${25^2} = {24^2} + {B^2}$
So,
$\Rightarrow$$625 = 576 + {B^2}$
On simplifying,
$\Rightarrow$$625 - 576 = {B^2}$
$\Rightarrow$$49 = {B^2}$
On solving we get ,
$\Rightarrow$Base = 7
Now we will find out the value of $\tan \theta $ and $\sec \theta $ .
\[\tan \theta = \dfrac{{perpendicular}}{{base}}\]
So, $\tan \theta = \dfrac{{24}}{7}$ but $\theta $lies in second quadrant and $\tan \theta $ is negative in second quadrant so $\tan \theta = - \dfrac{{24}}{7}$
$\sec \theta = \dfrac{{hypotenuse}}{{base}}$ but $\theta $lies in second quadrant and $\sec \theta $ is negative in second quadrant so
$\sec \theta = - \dfrac{{25}}{7}$
Now, we will find out value of $\sec \theta + \tan \theta $
$\sec \theta + \tan \theta = \left( { - \dfrac{{25}}{7}} \right) + \left( { - \dfrac{{24}}{7}} \right)$
$ \Rightarrow \dfrac{{ - 25 - 24}}{7}$
$ \Rightarrow \dfrac{{ - 49}}{7}$
$ \Rightarrow - 7$
So we have $\sec \theta + \tan \theta = - 7$

Hence, the correct option is option (C).

Note: As it is mentioned in the question that $\theta $ lies in the second quadrant, so it should not be ignored as in every quadrant different trigonometric ratios have different values. If this point is ignored then the whole solution will be changed and you will get a wrong answer as well.
Don’t forget properties and functions of different trigonometric ratios.