Question

What is the value of this Trigonometric expression?

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tan\left\{\dfrac{1}{2}\,\cos^{-1}\left(\dfrac{2}{\sqrt{5}}\right)\right\}$

$\bf{Let\,\,\,\theta=\dfrac{1}{2}\,cos^{-1}\left(\dfrac{2}{\sqrt{5}}\right)}$

$\sf{\implies\,2\theta=cos^{-1}\left(\dfrac{2}{\sqrt{5}}\right)}$

$\sf{\implies\,cos(2\theta)=\dfrac{2}{\sqrt{5}}}$

$\sf{\implies\,\dfrac{1-tan^2(\theta)}{1+tan^2(\theta)}=\dfrac{2}{\sqrt{5}}}$

$\sf{\implies\,\dfrac{1+tan^2(\theta)}{1-tan^2(\theta)}=\dfrac{\sqrt{5}}{2}$

Applying componendo and dividendo,

$\sf{\implies\,\dfrac{1+tan^2(\theta)+1-tan^2(\theta)}{1+tan^2(\theta)-1+tan^2(\theta)}=\dfrac{\sqrt{5}+2}{\sqrt{5}-2}$

$\sf{\implies\,\dfrac{2}{2\,tan^2(\theta)}=\dfrac{\sqrt{5}+2}{\sqrt{5}-2}$

$\sf{\implies\,tan(\theta)=\sqrt{\dfrac{\sqrt{5}-2}{\sqrt{5}+2}}$

$\sf{\implies\,tan(\theta)=\sqrt{\dfrac{\left(\sqrt{5}-2\right)\left(\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}$

$\sf{\implies\,tan(\theta)=\sqrt{\dfrac{\left(\sqrt{5}-2\right)^2}{\left(\sqrt{5}\right)^2-\left(2\right)^2}}$

$\sf{\implies\,tan(\theta)=\sqrt{\dfrac{\left(\sqrt{5}-2\right)^2}{5-4}}$

$\sf{\implies\,tan(\theta)=\sqrt{\dfrac{\left(\sqrt{5}-2\right)^2}{1}}$

$\sf{\implies\,tan(\theta)=\sqrt{5}-2}$

Now,

$\tan\left\{\dfrac{1}{2}\,\cos^{-1}\left(\dfrac{2}{\sqrt{5}}\right)\right\}=\tan(\theta)$

$\implies\tan\left\{\dfrac{1}{2}\,\cos^{-1}\left(\dfrac{2}{\sqrt{5}}\right)\right\}=\sqrt{5}-2$