Question

please solve this problem

Answer 1

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$given = \cos \alpha = \sin \alpha \\ i.e \: \cos45 = \sin45 = \frac{1}{2} \\ i.e \ \: alpha = 45 \\ \\ now \: (a) \: 2 {tan}^{2} \alpha + {cos}^{2} \alpha - 1 \\ 2 ({1})^{2} + \frac{1}{ \sqrt{2} } - 1 \\ 2 + \frac{1}{2} - 1 \\ 1 + \frac{1}{2} \\ \frac{3}{2} \\ \\ (b)2 {cos}^{2} \alpha - 1 \\ 2( \frac{1}{ \sqrt{2} } ) - 1 \\ \frac{2}{2} - 1 \\ 1 - 1 \\ 0$
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Answer 2

Step-by-step explanation:

We are given that ,

$\implies{\mathsf{ Cos(\theta) = Sin(\theta) }}$

Then it is clear that both sin and cos will be equal at 45° or Π/4 rad.

That is

$\implies{\mathsf{ Cos(\theta) = Sin(\theta) = 45^{\circ{}} = \frac{1}{\sqrt{2}} }}$

1]

Reduce the function —:

$\implies{\mathsf{ 2tan^{2}(\theta) + Cos^{2}(\theta) - 1 }}$

$\implies{\mathsf{ 2 \times ( \frac{ { \sin(\theta) }^{2} }{ { \cos(\theta) }^{2} } ) + { \cos(\theta) }^{2} - 1}} \\ \\ \implies{\mathsf{2 \times ( \frac{ { (\frac{1}{ \sqrt{2} } )}^{2} }{ { (\frac{1}{ \sqrt{2} } )}^{2} }) + { (\frac{1}{ \sqrt{2} }) }^{2} - 1 }} \\ \\ \implies{\mathsf{ 2( \frac{ \frac{1}{2} }{ \frac{1}{2} }) + \frac{1}{2} - 1 }} \\ \\ \implies{\mathsf{ \frac{1}{2} + 1 }} \\ \\ \implies{\mathsf{ \frac{3}{2}}}$

2]

$\implies{\mathsf{ 2Cos^{2}(\theta) - 1 }}$

$\implies{\mathsf{2 \times ( { \frac{1}{ \sqrt{2} } )}^{2} - 1 }} \\ \\ \implies{\mathsf{2 \times ( \frac{1}{2} ) - 1 }} \\ \\ \implies{\mathsf{1 - 1 }}\\ \\ \implies{\mathsf{= 0}}$