Question

If sec theta + tan theta = 2/3, find the value of sin theta and determine the quadrant in which theta lies.

Answer 1

Step-by-step explanation:

IN THE PICTURE...

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PLZ MARK BRAINLIEST

Answer 2

The value of  $\text{sin} \theta =\frac{-5}{13}$ and theta lies in the 3rd and 4th quadrant.

Step-by-step explanation:

We are given that sec theta + tan theta = 2/3 and we have to determine the value of sin theta.

The given equation is;

$\text{sec} \theta + \text{tan} \theta = \frac{2}{3}$

As we know that  $\text{sec} \theta = \frac{1}{\text{cos} \theta}$  and  $\text{tan} \theta = \frac{\text{sin} \theta}{\text{cos} \theta}$ , so;

$\frac{1}{\text{cos} \theta} + \frac{\text{sin} \theta}{\text{cos} \theta} = \frac{2}{3}$

$\frac{1+\text{sin} \theta}{\text{cos} \theta} = \frac{2}{3}$

Now, squaring both sides we get;

$(\frac{1+\text{sin} \theta}{\text{cos} \theta} )^{2} = (\frac{2}{3})^{2}$

$\frac{(1+\text{sin} \theta)^{2} }{\text{cos}^{2} \theta} = \frac{4}{9}$

$\frac{(1+\text{sin} \theta)\times (1+\text{sin} \theta) }{1-\text{sin}^{2} \theta} = \frac{4}{9}$           {$\because \text{sin}^{2} \theta+ \text{cos}^{2} \theta= 1$}

$\frac{(1+\text{sin} \theta)\times (1+\text{sin} \theta) }{(1-\text{sin} \theta)(1+\text{sin} \theta}) = \frac{4}{9}$          {$a^{2}-b^{2} = (a-b)(a+b)$}

$\frac{1+\text{sin} \theta }{1-\text{sin} \theta} = \frac{4}{9}$

Now, cross multiplying the fractions we get;

$9 \times (1+\text{sin} \theta) = 4 \times (1-\text{sin} \theta)$

$9+9\text{sin} \theta = 4-4\text{sin} \theta$

$9\text{sin} \theta+4\text{sin} \theta = 4-9$

$13\text{sin} \theta = -5$

$\text{sin} \theta =\frac{-5}{13}$

Since the value of $\text{sin} \theta$ is negative, this means that theta will lie in 3rd and 4th quadrant because in these two-quadrants sin values are negative.