Question

If $0<\theta <\frac{\pi }{2}$ and $\sin \theta +\cos \theta +\tan \theta +\cot \theta +\sec \theta +csc\theta$ is equal to $7$, then $\sin 2\theta$ is a root of the equation

• A. $x^2+44x+36=0$
• B. $x^2-44x-36=0$
• C. $x^2-44x+36=0$
• D. $x^2+44x-36=0$

Answer 1

The correct option is C $x^2-44x+36=0$

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