If $0 < θ < \frac{π}{2}$ and $\sin θ + \cos θ + \tan θ + \cot θ + \sec θ + \cos e c θ$ is equal to 7, then $\sin 2 θ$ is a root of the equation

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$x^{2} + 44 x + 36 = 0$

$x^{2} - 44 x - 36 = 0$

$x^{2} - 44 x + 36 = 0$

$x^{2} + 44 x - 36 = 0$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\sin θ + \cos θ + \tan θ + \cot θ + \sec θ + cosec θ = 7$
$\Rightarrow \sin θ + \cos θ + \frac{\sin θ}{\cos θ} + \frac{\cos θ}{\sin θ} + \frac{1}{\cos θ} + \frac{1}{\sin θ} = 7$
$\Rightarrow \frac{\sin^{2} θ \cos θ + \cos^{2} θ \sin θ + \sin^{2} θ + \cos^{2} θ + \sin θ}{\sin θ \cos θ}$
$\Rightarrow \frac{\sin θ \cos θ (\sin θ + \cos θ) + 1 + \sin θ + \cos θ}{\sin θ \cos θ} = 7$
Substituting $2 \sin θ \cos θ = x \Rightarrow \sin θ + \cos θ = \sqrt{1 + x}$
We get $\frac{\frac{x}{2} (\sqrt{1 + x}) + 1 + \sqrt{1 + x}}{\frac{x}{2}} = 7 \Rightarrow \sqrt{1 + x} (x + 2) + 2 = 7 x$
Squaring both sides, we get
$(1 + x) {(x + 2)}^{2} = {(7 x - 2)}^{2}$
$\Rightarrow (1 + x) (x^{2} + 4 + 4 x) = 49 x^{2} + 4 - 28 x$
$\Rightarrow x^{2} + 4 + 4 x + x^{3} + 4 x + 4 x^{2} = 49 x^{2} + 4 - 28 x$
$\Rightarrow x^{3} - 44 x^{2} + 36 x = 0 \Rightarrow x^{2} - 44 x + 36 = 0$
Hence, option 'C' is correct.