Question

If x=5+2root6 and cos theta=(2rootx)/(x-1) , then cosec^2theta+8=??

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\blue{x=5+2\sqrt{6}\,\,\,\,and\,\,\,\,cos(\theta)=\dfrac{2\sqrt{x}}{x-1}}}$

Now,

$\sf{cos(\theta)=\dfrac{2\sqrt{5+2\sqrt{6} }}{5+2\sqrt{6}-1}}$

$\sf{\implies\,cos(\theta)=\dfrac{2\sqrt{3+2+2\sqrt{6} }}{4+2\sqrt{6}}}$

$\sf{\implies\,cos(\theta)=\dfrac{2\sqrt{(\sqrt{3})^2 +(\sqrt{2})^2+2\cdot\sqrt{3}\cdot\sqrt{2} }}{2\sqrt{2}(\sqrt{2}+\sqrt{3})}}$

$\sf{\implies\,cos(\theta)=\dfrac{\sqrt{(\sqrt{3}+\sqrt{2})^2 }}{\sqrt{2}(\sqrt{2}+\sqrt{3})}}$

$\sf{\implies\,cos(\theta)=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{2}(\sqrt{2}+\sqrt{3})}}$

$\sf{\implies\,cos(\theta)=\dfrac{1}{\sqrt{2}}}$

So,

$\sf{cosec^2(\theta)+8}$

$\sf{=\dfrac{1}{sin^2(\theta)}+8}$

$\sf{=\dfrac{1}{1-cos^2(\theta)}+8}$

$\sf{=\dfrac{1}{1-\bigg(\dfrac{1}{\sqrt{2} }\bigg)^2}+8}$

$\sf{=\dfrac{1}{1-\dfrac{1}{2 }}+8}$

$\sf{=\dfrac{1}{\dfrac{1}{2 }}+8}$

$\sf{=2+8}$

$\sf{=10}$