What is the value of 4cot^3(135)+7tan^4(150)-csc^2(240) ?

The value of 4cot^3(135)+7tan^4(150)-csc^2(240) is -41/9

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4cot^3(135)+7tan^4(150)-csc^2(240)

4 cot^{3} (13 5^{\circ}) + 7 tan^{4} (15 0^{\circ}) - csc^{2} (24 0^{\circ})

- \frac{41}{9}

Solution steps

4 cot^{3} (13 5^{\circ}) + 7 tan^{4} (15 0^{\circ}) - csc^{2} (24 0^{\circ})

Rewrite using trig identities:

csc^{2} (24 0^{\circ}) = 1 + cot^{2} (6 0^{\circ})

= 4 cot^{3} (13 5^{\circ}) + 7 tan^{4} (15 0^{\circ}) - (1 + cot^{2} (6 0^{\circ}))

Simplify

= 4 cot^{3} (13 5^{\circ}) + 7 tan^{4} (15 0^{\circ}) - 1 - cot^{2} (6 0^{\circ})

Use the following trivial identity:

cot (13 5^{\circ}) = - 1

Rewrite using trig identities:

tan (15 0^{\circ}) = - \frac{3}{3}

Use the following trivial identity:

cot (6 0^{\circ}) = \frac{3}{3}

= 4 (- 1)^{3} + 7 (- \frac{3}{3})^{4} - 1 - (\frac{3}{3})^{2}

Simplify

4 (- 1)^{3} + 7 (- \frac{3}{3})^{4} - 1 - (\frac{3}{3})^{2} : - \frac{41}{9}

= - \frac{41}{9}

cos (π (2))

arccos (\frac{3.9}{8.5})

\frac{tan ( 1 2 ^{\circ} )}{1 - tan ^{2} ( 1 2 ^{\circ} )}

sin (10 5^{\circ}) sin (1 5^{\circ})

sec^{2} (- π)

What is the value of 4cot^3(135)+7tan^4(150)-csc^2(240) ?
The value of 4cot^3(135)+7tan^4(150)-csc^2(240) is -41/9