Question

If $1+co{t}^2\theta ={(\sqrt{3+2\sqrt{2}}-1)}^2$, find the value of $\frac{1}{tan\theta }+\frac{sin\theta }{1+cos\theta }$

• A. $-2$
• B. $1+\sqrt{2}$
• C. $\sqrt{2}$
• D. $1$

Answer 1

The correct option is C $\sqrt{2}$

$1+co{t}^2\theta =cose{c}^2\theta ={(\sqrt{3+2\sqrt{2}}-1)}^2;$
On comparing, $cosec\theta =\sqrt{3+2\sqrt{2}}-1$
$\sqrt{3+2\sqrt{2}}$ can be written as
$\sqrt{1+2+2\sqrt{2}}=\sqrt{(1{)}^2+(\sqrt{2}{)}^2+2\times 1\times \sqrt{2}}$
On comparing with $(a^2+b^2+2ab)=(a+b{)}^2$
We can write as,
$\sqrt{(1+\sqrt{2}{)}^2}=1+\sqrt{2}$
So, cosec$\theta =1+\sqrt{2}-1$
cosec$\theta =\sqrt{2}=$ cosec ${45}^{\circ }$
$\theta ={45}^{\circ }$
$\frac{1}{tan\theta }+\frac{sin\theta }{1+cos\theta }=\frac{1}{tan{45}^{\circ }}+\frac{sin{45}^{\circ }}{1+cos{45}^{\circ }}$
$=\frac{1}{1}+\frac{\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}$
$=1+\frac{1}{1+\sqrt{2}}$
On simplifying we get, $\sqrt{2}$