4sin 430-cos 460-2 - 3sin 2 60-cos 2 45-1 - 2 tan 2 60-

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Standard X

Mathematics

Standard Values of Trigonometric Ratios

4sin 430+cos ...

Question

4(sin⁴30+cos⁴60)-2/3(sin²60-cos²45)+1/2tan²60

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Solution

Substitute the values of trignometric functions to get the answer
sin30 = 1/2 cos 60 = 1/2 sin60 = √3/2 cos 45 = 1/√2 tan60 = √3
substituting the values in the equation, 4(sin⁴30+cos⁴60)-2/3(sin²60-cos²45)+1/2tan²60
= 4( 1/16 + 1/16) - 2/3( 3/4 - 1/2) + 1/2x3
= 4/8 - 1/6 + 3/2
= 1/2 - 1/6 + 3/2
= 11/6

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Similar questions

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4 ({sin}^{4} 30^{\circ} + {cos}^{4} 60^{\circ}) - \frac{2}{3} ({sin}^{2} 60^{\circ} - {cos}^{2} 45^{\circ}) + \frac{1}{2} {tan}^{2} 60^{\circ}

Q. The value of

4 ({sin}^{4} 30^{\circ} + c o s^{4} 60^{\circ}) ({cos}^{2} 45^{\circ} - {sin}^{2} 90^{\circ}) + {tan}^{4} 45^{\circ} - {cot}^{2} 45^{\circ}

4 (\sin^{4} 30 ° + \cos^{4} 60 °) - \frac{2}{3} (\sin^{2} 60 ° - \cos^{2} 45 °) + \frac{1}{2} \tan^{2} 60 ° = ?

(a)

\frac{8}{3}

(b)

\frac{11}{6}

(c)

\frac{13}{6}

(d)

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(i) $\frac{2}{3} (c o s^{4} 30^{o} - s i n^{4} 45^{o}) - 3 (s i n^{2} 60^{o} - s e c^{2} 45^{o}) + \frac{1}{4} c o t^{2} 30^{o}$

(ii) $4 (s i n^{4} 30^{o} + c o s^{4} 60^{o}) - \frac{2}{3} (s i n^{2} 60^{o} - c o s^{2} 45^{o}) + \frac{1}{2} t a n^{2} 60^{o}$

(iii) $\frac{s i n 50^{o}}{c o s 40^{o}} + \frac{c o s e c 40^{o}}{s e c 50^{o}} - 4 c o s 50^{o} c o s e c 40^{o}$

(iv) $t a n 35^{o} t a n 40^{o} t a n 45^{o} t a n 50^{o} t a n 55^{o}$

(v) $c o s e c (65^{o} + θ) - s e c (25^{o} - θ) - t a n (55^{o} - θ) + c o t (35^{o} + θ)$

(vi) $t a n 7^{o} t a n 23^{o} t a n 60^{o} t a n 67^{o} t a n 83^{o}$

(vii) $\frac{2 s i n 68^{o}}{c o s 22^{o}} - \frac{2 c o t 15^{o}}{5 t a n 75^{o}} - \frac{3 t a n 45^{o} t a n 20^{o} t a n 40^{o} t a n 50^{o} t a n 70^{o}}{5}$
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(ix) $\frac{s i n 18^{o}}{c o s 72^{o}} + \sqrt{3} {t a n 10^{o} t a n 30^{o} t a n 40^{o} t a n 50^{o} t a n 80^{o}}$
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Q. Evaluate

4 ({sin}^{4} 30^{\circ} + {cos}^{4} 60^{\circ} - 3 ({cos}^{2} 45^{\circ} - {sin}^{2} 90^{\circ})

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