What is the value of 1+tan^2(60)-sec^2(60) ?

The value of 1+tan^2(60)-sec^2(60) is 0

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1+tan^2(60)-sec^2(60)

1 + tan^{2} (6 0^{\circ}) - sec^{2} (6 0^{\circ})

0

Solution steps

1 + tan^{2} (6 0^{\circ}) - sec^{2} (6 0^{\circ})

Use the following trivial identity:

tan (6 0^{\circ}) = 3

tan (6 0^{\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}

= 3

Use the following trivial identity:

sec (6 0^{\circ}) = 2

sec (6 0^{\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

= 1 + (3)^{2} - 2^{2}

Simplify

1 + (3)^{2} - 2^{2} : 0

1 + (3)^{2} - 2^{2}

(3)^{2} = 3

(3)^{2}

Apply radical rule:

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

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= 3^{\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

= 3

= 1 + 3 - 2^{2}

Add the numbers:

1 + 3 = 4

= 4 - 2^{2}

2^{2} = 4

= 4 - 4

Subtract the numbers:

4 - 4 = 0

= 0

= 0

cos (\frac{π}{4}) - \frac{π}{4}

(\frac{3}{2}) (\frac{9.8}{0.4}) (1 - sin (6 0^{\circ}) + cos (6 0^{\circ}))

cos (0.68)

cos (15 0^{\circ}) - sin (24 0^{\circ}) + tan (30 0^{\circ})

arctan (\frac{- \frac{3}{2}}{\frac{1}{2}})

What is the value of 1+tan^2(60)-sec^2(60) ?
The value of 1+tan^2(60)-sec^2(60) is 0