If sectheta+tantheta=3/2, what is the value of sintheta?

Dec 12, 2016

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

Explanation:

$\tan \theta + \sec \theta = \frac{3}{2}$
No idea-well do something!!
$\sin \frac{\theta}{\cos} \theta + \frac{1}{\cos} \theta = \frac{3}{2}$
And then the next stage becomes clear
$\frac{\sin \theta + 1}{\cos} \theta = \frac{3}{2}$
$2 \left(\sin \theta + 1\right) = 3 \cos \theta$

So what now.? Well we only want to know about $\sin \theta$
And we do know ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$
So square both sides

$4 {\left(\sin \theta + 1\right)}^{2} = 9 {\cos}^{2} \theta$
$4 \left({\sin}^{2} \theta + 2 \sin \theta + 1\right) = 9 \left(1 - {\sin}^{2} \theta\right)$
$4 {\sin}^{2} \theta + 8 \sin \theta + 4 = 9 - 9 {\sin}^{2} \theta$
$13 {\sin}^{2} \theta + 8 \sin \theta - 5 = 0$
Factorise
$\left(13 \sin \theta - 5\right) \left(\sin \theta + 1\right) = 0$
$\sin \theta = \frac{5}{13}$ or $\sin \theta = - 1$
Only the first will do because if $\sin \theta = - 1$ then $\cos \theta = 0$ and clearly we cannot divide by zero.

Dec 12, 2016

Given

$\sec \theta + \tan \theta = 1.5 = \frac{15}{10} = \frac{3}{2}$

=>sectheta+tantheta=3/2……………(1)

Again we know

sec^2theta-tan^2theta=1……………(2)

Dividing (2) by (1) we get

sectheta-tantheta=2/3……………(3)

Adding (1) and (3) we get

$2 \sec \theta = \frac{3}{2} + \frac{2}{3} = \frac{13}{6}$

$\implies \sec \theta = \frac{13}{12}$

Subtracting (3) from (1) we get

$2 \tan \theta = \frac{3}{2} - \frac{2}{3} = \frac{5}{6}$

$\implies \tan \theta = \frac{5}{12}$

$\implies \sin \theta \times \sec \theta = \frac{5}{12}$

$\implies \sin \theta \times \frac{13}{12} = \frac{5}{12}$

$\implies \sin \theta = \frac{5}{12} \times \frac{12}{13} = \frac{5}{13}$

Alternative

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

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

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

$\implies \frac{1 + \sin \theta}{\sqrt{1 - {\sin}^{2} \theta}} = \frac{3}{2}$

$\implies \frac{\sqrt{1 + \sin \theta} \sqrt{1 + \sin \theta}}{\sqrt{1 - \sin \theta} \sqrt{1 + \sin \theta}} = \frac{3}{2}$

for $\sin \theta \ne - 1$

$\implies \frac{\sqrt{1 + \sin \theta}}{\sqrt{1 - \sin \theta}} = \frac{3}{2}$

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

$\implies 4 + 4 \sin \theta = 9 - 9 \sin \theta$

$\implies 13 \sin \theta = 5$

$\implies \sin \theta = \frac{5}{13}$