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Trigonometric Ratio(sin,cos,tan)

(3+1)(3-cot30...

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(3+1)(3-cot30)=tanpower3 60^° -2sin60^°

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Q. Prove the following:

(\sqrt{3} + 1) (3 - cot 30^{\circ}) = {tan}^{3} 60^{\circ} - 2 sin 60^{\circ}

Q. Show that (√3+1)(3-cot30) is equal to tan cube 60 minus 2 sin 60

(\sqrt{3} + 1) (3 - c o t 30^{\circ}) = t a n^{3} 60^{\circ} - 2 s i n 60^{\circ}

(\sqrt{3} + 1) (3 - c o t 30^{°}) = \tan^{3} 60^{°} - 2 \sin 60^{°}

\sqrt{2}

sin (60° – α) = 1 then α = ?
(a) 15°
(b) 30°
(c) 45°
(d) 60°

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