What is the value of 5tan^2(45)+2sec^2(45) ?

The value of 5tan^2(45)+2sec^2(45) is 9

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5tan^2(45)+2sec^2(45)

5 tan^{2} (4 5^{\circ}) + 2 sec^{2} (4 5^{\circ})

9

Solution steps

5 tan^{2} (4 5^{\circ}) + 2 sec^{2} (4 5^{\circ})

Use the following trivial identity:

tan (4 5^{\circ}) = 1

tan (4 5^{\circ})

tan (x)

periodicity table with

18 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 1

Use the following trivial identity:

sec (4 5^{\circ}) = 2

sec (4 5^{\circ})

sec (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

= 2

= 5 \cdot 1^{2} + 2 (2)^{2}

Simplify

5 \cdot 1^{2} + 2 (2)^{2} : 9

5 \cdot 1^{2} + 2 (2)^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= 5 \cdot 1 + 2 (2)^{2}

5 \cdot 1 = 5

5 \cdot 1

Multiply the numbers:

5 \cdot 1 = 5

= 5

2 (2)^{2} = 4

2 (2)^{2}

(2)^{2} = 2

(2)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 2

= 2 \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 4

= 5 + 4

Add the numbers:

5 + 4 = 9

= 9

= 9

4 - sin (3)

csc (7 4^{\circ})

1 cos (\frac{π}{2})

cos (2.4)

csc (\frac{- 9 π}{4})

What is the value of 5tan^2(45)+2sec^2(45) ?
The value of 5tan^2(45)+2sec^2(45) is 9