Question

4sin460° + 4cos4 30° - 3(tan-60° - tan245°) +
5cos245º = 1​

Answer 1

$4 sin^{2}60 + 4cos^{2}30 - 3(tan^{2}60 - tan^{2}45) + 5cos^45$

= $4(\frac{\sqrt{3} }{2} )^4 + 4(\frac{\sqrt{3} }{2} )^4 - 3[(\sqrt{3} )^2- (1)^2] - 5(\frac{1}{\sqrt{2} } )^2$

= $8(\frac{9}{16}) - 3(3-1) + 5(\frac{1}{2})$

= $\frac{9}{2} - 6 + \frac{5}{2}$

= $\frac{9-12+5}{2}$

= $\frac{2}{2}$

= 1

∴ LHS = RHS

Substitute the trigonometric ratios in the given equation and solve it like an algebraic equation.

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