# Question #39bdb

Oct 1, 2017

43/20

#### Explanation:

Given, $9 {\cot}^{2} \theta = 18 \cos e c \theta - 14$

$\Rightarrow 9 \left(\cos e {c}^{2} \theta - 1\right) - 18 \cos e c \theta + 14 = 0$

$\Rightarrow 9 \cos e {c}^{2} \theta - 9 - 18 \cos e c \theta + 14 = 0$

$\Rightarrow 9 \cos e {c}^{2} \theta - 18 \cos e c \theta + 5 = 0$

$\Rightarrow 9 \cos e {c}^{2} \theta - 15 \cos e c \theta - 3 \cos e c \theta + 5 = 0$

$\Rightarrow 3 \cos e c \theta \left(3 \cos e c \theta - 5\right) - 1 \left(3 \cos e c \theta - 5\right)$

$\Rightarrow \left(3 \cos e c \theta - 5\right) \left(3 \cos e c \theta - 1\right) = 0$

$\Rightarrow 3 \cos e c \theta - 5 = 0 , 3 \cos e c \theta - 1 = 0$

$\Rightarrow \cos e c \theta = \frac{5}{3} \mathmr{and} \frac{1}{3}$ [ cosectheta can not be 1/3 as theta should be between 0 < theta< pi/2]

Hence $\cos e c \theta = \frac{5}{3} \mathmr{and} \sin \theta = \frac{3}{5} , \cos \theta = \frac{4}{5} \mathmr{and} \tan \theta = \frac{3}{4}$

S0, $\sin \theta + \cos \theta + \tan \theta = \frac{3}{5} + \frac{4}{5} + \frac{3}{4} = \frac{43}{20}$