Find the Exact Value 12tan(pi/4)^2+24sec(pi/4)^2

$12 \tan^{2} (\frac{π}{4}) + 24 \sec^{2} (\frac{π}{4})$

Simplify each term.

The exact value of $\tan (\frac{π}{4})$ is $1$ .

$12 \cdot 1^{2} + 24 \sec^{2} (\frac{π}{4})$

One to any power is one.

$12 \cdot 1 + 24 \sec^{2} (\frac{π}{4})$

Multiply $12$ by $1$ .

$12 + 24 \sec^{2} (\frac{π}{4})$

The exact value of $\sec (\frac{π}{4})$ is $\frac{2}{\sqrt{2}}$ .

$12 + 24 {(\frac{2}{\sqrt{2}})}^{2}$

Multiply $\frac{2}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$12 + 24 {(\frac{2}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}})}^{2}$

Combine and simplify the denominator.

Multiply $\frac{2}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$12 + 24 {(\frac{2 \sqrt{2}}{\sqrt{2} \sqrt{2}})}^{2}$

Raise $\sqrt{2}$ to the power of $1$ .

$12 + 24 {(\frac{2 \sqrt{2}}{{\sqrt{2}}^{1} \sqrt{2}})}^{2}$

Raise $\sqrt{2}$ to the power of $1$ .

$12 + 24 {(\frac{2 \sqrt{2}}{{\sqrt{2}}^{1} {\sqrt{2}}^{1}})}^{2}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$12 + 24 {(\frac{2 \sqrt{2}}{{\sqrt{2}}^{1 + 1}})}^{2}$

Add $1$ and $1$ .

$12 + 24 {(\frac{2 \sqrt{2}}{{\sqrt{2}}^{2}})}^{2}$

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$12 + 24 {(\frac{2 \sqrt{2}}{{(2^{\frac{1}{2}})}^{2}})}^{2}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$12 + 24 {(\frac{2 \sqrt{2}}{2^{\frac{1}{2} \cdot 2}})}^{2}$

Combine $\frac{1}{2}$ and $2$ .

$12 + 24 {(\frac{2 \sqrt{2}}{2^{\frac{2}{2}}})}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$12 + 24 {(\frac{2 \sqrt{2}}{2^{\frac{2}{2}}})}^{2}$

Rewrite the expression.

$12 + 24 {(\frac{2 \sqrt{2}}{2^{1}})}^{2}$

Evaluate the exponent.

$12 + 24 {(\frac{2 \sqrt{2}}{2})}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$12 + 24 {(\frac{2 \sqrt{2}}{2})}^{2}$

Divide $\sqrt{2}$ by $1$ .

$12 + 24 {\sqrt{2}}^{2}$

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$12 + 24 {(2^{\frac{1}{2}})}^{2}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$12 + 24 \cdot 2^{\frac{1}{2} \cdot 2}$

Combine $\frac{1}{2}$ and $2$ .

$12 + 24 \cdot 2^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$12 + 24 \cdot 2^{\frac{2}{2}}$

Rewrite the expression.

$12 + 24 \cdot 2^{1}$

Evaluate the exponent.

$12 + 24 \cdot 2$

Multiply $24$ by $2$ .

$12 + 48$

Add $12$ and $48$ .

$60$

Find the Exact Value 12tan(pi/4)^2+24sec(pi/4)^2

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