Question

10) If alpha= 45° then find the value of

$\frac{1+{sin}^{2} \alpha }{ {cos}^{2} \alpha }$


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Answer 1

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${ \sin( \alpha ) }^{2} = \frac{1}{2}$

${ \cos( \alpha ) }^{2} = \frac{1}{2}$

= 3.

Hope this helps you.....

Answer 2

Question :-

$10) \: \sf if \: \alpha = 45 \degree \: then \: find - \\ \to \: \frac{1 + { \sin}^{2} \alpha }{ { \cos}^{2} \alpha}$

Answer :-

$\implies \: \boxed{\: \frac{1 + { \sin}^{2} \alpha }{ { \cos}^{2} \alpha} = 3}$

Explanation :-

we have ,

$\to \: \frac{1 + { \sin}^{2} \alpha }{ { \cos}^{2} \alpha} \\ \: \\ \sf \: \: put \: \alpha = 45 \degree \\ \\ \to \: \frac{1 + { \sin}^{2}45 \degree }{ { \cos}^{2} 45 \degree } \\ \\ \because \: \sin45 \degree = \frac{1}{ \sqrt{2} } = \cos45 \degree \\ \\ \to \: \frac{1 + { (\frac{1}{ \sqrt{2} }) }^{2} }{ { (\frac{1}{ \sqrt{2} } )}^{2} } \\ \\ \to \: \frac{1 + \frac{1}{2} }{ \frac{1}{2} } \\ \\ \to \: \frac{ \frac{3}{2} }{ \frac{1}{2} } \\ \\ \to \: 3 \\ \\ \implies \: \boxed{\: \frac{1 + { \sin}^{2} \alpha }{ { \cos}^{2} \alpha} = 3}$