Question

If 2cos^2theta -1 =0 and angle theta is less than 90 degree,find the value of theta and hence find the value of sin^2theta + tan^2 theta​

Answer 1

$\huge\sf{Answer:}$

$\large\sf{\implies\:\theta=45\degree}$

$\large\sf{\implies\:{\sin}^{2}\theta +{\tan}^{2}\theta=\frac{3}{2}}$

$\large\sf{\underline{\star{Step-by-Step-Explanation-}}}$

$\large\sf{\implies 2 \: { \cos}^{2} \theta - 1 = 0 }$

$\large\sf{\implies 2 \: { \cos}^{2} \theta = 1 }$

$\large\sf{ \implies { \cos}^{2} \theta = \frac{1}{2} }$

$\large\sf{\implies \cos \: \theta = \sqrt{ \frac{1}{2} }}$

$\large\sf{\implies \cos \: \theta = \frac{1}{ \sqrt{2} } }$

$\large\sf{\implies \cos \: \theta = \cos \: 45 \degree }$

$\large\sf{\implies \theta = 45 \degree}$

$\large\sf{Now,}$

$\large\sf{={\sin}^{2}\theta +{\tan}^{2}\theta}$

$\large\sf{={\sin}^{2}\:45\degree+{\tan}^{2}\:45\degree }$

$\large\sf{ = {(\frac{1}{\sqrt{2}})}^{2}+{(1)}^{2} }$

$\large\sf{=\frac{1}{2}+1}$

$\large\sf{=\frac{1+2}{2}}$

$\large\sf{=\frac{3}{2}}$

$\large\sf{Hence,}$

$\large\sf{\boxed{\implies\:\theta=45\degree}}$

$\large\sf{\boxed{\implies\:{\sin}^{2}\theta +{\tan}^{2}\theta=\frac{3}{2}}}$