Question

If theta is an acute angle and sin theta = cos theta , find the value of 2 sin square theta - 3 cos square theta + 1/2 cot square theta.

please solve this problem if you can't understand the question you can see the image i have given and please solve this step by step.​

Answer 1

Answer:

0

Step-by-step explanation:

For simplicity, just assume the angle theta as angle alpha.

Given that,

$\sin( \alpha ) = \cos( \alpha )$

Solving this, we will get,

$= > \frac{ \sin( \alpha ) }{ \cos( \alpha ) } = 1 \\ \\ = > \tan( \alpha ) = 1 \\ \\ = > \tan( \alpha ) = \tan( \frac{\pi}{4} ) \\ \\ = > \alpha = \frac{\pi}{4}$

Now, to find the value of,

$2 { \sin }^{2} \alpha - 3 { \cos }^{2} \alpha + \frac{1}{2} { \cot }^{2} \alpha$

Simplifying this, we get,

$2 { \sin }^{2} \frac{\pi}{4} - 3 { \cos }^{2} \frac{\pi}{4} + \frac{1}{2} { \cot }^{2} \frac{\pi}{4}$

Now, we know that,

• $\sin( \frac{\pi}{4} ) = \frac{1}{ \sqrt{2} }$
• $\cos( \frac{\pi}{4} ) = \frac{1}{ \sqrt{2} }$
• $\cot( \frac{\pi}{4} ) = 1$

Therefore, substituting the values, we get,

$= 2 {( \frac{1}{ \sqrt{2} }) }^{2} - 3 {( \frac{1}{ \sqrt{2} }) }^{2} + \frac{1}{2} {(1)}^{2} \\ \\ = 2 \times \frac{1}{2} - 3 \times \frac{1}{2} + \frac{1}{2} \times 1 \\ \\ = 1 - \frac{3}{2} + \frac{1}{2} \\ \\ = \frac{2 - 3 + 1}{2} \\ \\ = \frac{3 - 3}{2} \\ \\ = \frac{0}{2} \\ \\ = 0$

Hence, required value is 0.

Answer 2

Answer:

0

Step-by-step explanation:

For simplicity, just assume the angle theta as angle alpha.

Given that,

\sin( \alpha ) = \cos( \alpha )sin(α)=cos(α)

Solving this, we will get,

\begin{gathered} = > \frac{ \sin( \alpha ) }{ \cos( \alpha ) } = 1 \\ \\ = > \tan( \alpha ) = 1 \\ \\ = > \tan( \alpha ) = \tan( \frac{\pi}{4} ) \\ \\ = > \alpha = \frac{\pi}{4} \end{gathered}

=>

cos(α)

sin(α)

=1

=>tan(α)=1

=>tan(α)=tan(

4

π

)

=>α=

4

π

Now, to find the value of,

2 { \sin }^{2} \alpha - 3 { \cos }^{2} \alpha + \frac{1}{2} { \cot }^{2} \alpha2sin

2

α−3cos

2

α+

2

1

cot

2

α

Simplifying this, we get,

2 { \sin }^{2} \frac{\pi}{4} - 3 { \cos }^{2} \frac{\pi}{4} + \frac{1}{2} { \cot }^{2} \frac{\pi}{4}2sin

2

4

π

−3cos

2

4

π

+

2

1

cot

2

4

π

Now, we know that,

\sin( \frac{\pi}{4} ) = \frac{1}{ \sqrt{2} }sin(

4

π

)=

2

1

\cos( \frac{\pi}{4} ) = \frac{1}{ \sqrt{2} }cos(

4

π

)=

2

1

\cot( \frac{\pi}{4} ) = 1cot(

4

π

)=1

Therefore, substituting the values, we get,

\begin{gathered} = 2 {( \frac{1}{ \sqrt{2} }) }^{2} - 3 {( \frac{1}{ \sqrt{2} }) }^{2} + \frac{1}{2} {(1)}^{2} \\ \\ = 2 \times \frac{1}{2} - 3 \times \frac{1}{2} + \frac{1}{2} \times 1 \\ \\ = 1 - \frac{3}{2} + \frac{1}{2} \\ \\ = \frac{2 - 3 + 1}{2} \\ \\ = \frac{3 - 3}{2} \\ \\ = \frac{0}{2} \\ \\ = 0\end{gathered}

=2(

2

1

)

2

−3(

2

1

)

2

+

2

1

(1)

2

=2×

2

1

−3×

2

1

+

2

1

×1

=1−

2

3

+

2

1

=

2

2−3+1

=

2

3−3

=

2

0

=0

hence your anger is 0