23 cos 430-sin 445-3 sin 260- 245-14 230- -a 5312b 7324c 11324d 8312

(c) nbsp;11324 23cos4 #160;30o sin4 #160;45o 3sin2 #160;60o sec2 #160;45o+14cot2 #160;30o=23324 124 3322 22+14 #215;32=23916 14 334 2+34=23 #2 ...

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Byju's Answer

Standard XII

Mathematics

Factorization Method Form to Remove Indeterminate Form

23 cos 430∘-s...

Question

$\frac{2}{3} (\cos^{4} 30 ° - \sin^{4} 45 °) - 3 (\sin^{2} 60 ° - \sec^{2} 45 °) + \frac{1}{4} \cot^{2} 30 ° = ?$
(a) $\frac{53}{12}$
(b) $\frac{73}{24}$
(c) $\frac{113}{24}$
(d) $\frac{83}{12}$

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Solution

(c) $\frac{113}{24}$

$\frac{2}{3} (\cos^{4} 30^{o} - \sin^{4} 45^{o}) - 3 (\sin^{2} 60^{o} - \sec^{2} 45^{o}) + \frac{1}{4} \cot^{2} 30^{o} = \frac{2}{3} [{(\frac{\sqrt{3}}{2})}^{4} - {(\frac{1}{\sqrt{2}})}^{4}] - 3 [{(\frac{\sqrt{3}}{2})}^{2} - {(\sqrt{2})}^{2}] + \frac{1}{4} \times {(\sqrt{3})}^{2} = \frac{2}{3} (\frac{9}{16} - \frac{1}{4}) - 3 (\frac{3}{4} - 2) + \frac{3}{4} = \frac{2}{3} \times \frac{5}{16} - 3 \times (- \frac{5}{4}) + \frac{3}{4} = \frac{5}{24} + \frac{15}{4} + \frac{3}{4} = \frac{5 + 90 + 18}{24} = \frac{113}{24}$

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Similar questions

Q. Evaluate :

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

Q. Evaluate :

(i)

\frac{2}{3} (\cos^{4} 30 - \sin^{4} 45 °) - 3 (\sin^{2} 60 ° - \sec^{2} 45 °) + \frac{1}{4} \cot^{2} 30 °

(ii)

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

(iii)

\frac{\sin 50 °}{\cos 40 °} + \frac{cosec 40 °}{\sec 50 °} - 4 \cos 50 ° cosec 40 °

(iv) tan 35° tan 40° tan 45° tan 50° tan 55°
(v) cosec (65° + θ) − sec (25° − θ) − tan (55° − θ) + cot (35° + θ)
(vi) tan 7° tan 23° tan 60° tan 67° tan 83°
(vii)

\frac{2 \sin 68 °}{\cos 22 °} - \frac{2 \cot 15 °}{5 \tan 75 °} - \frac{3 \tan 45 ° \tan 20 ° \tan 40 ° \tan 50 ° \tan 70 °}{5}

(viii)

\frac{3 \cos 55 °}{7 \sin 35 °} - \frac{4 (\cos 70 ° cosec 20 °)}{7 (\tan 5 ° \tan 25 ° \tan 45 ° \tan 65 ° \tan 85 °)}

(ix)

\frac{\sin 18 °}{\cos 72 °} + \sqrt{3} \{\tan 10 ° \tan 30 ° \tan 40 ° \tan 50 ° \tan 80 °\}

(x)

\frac{\cos 58 °}{\sin 32 °} + \frac{\sin 22 °}{\cos 68 °} - \frac{\cos 38 ° cosec 52 °}{\tan 18 ° \tan 35 ° \tan 60 ° \tan 72 ° \tan 55 °}

Evaluate : 2/3(cos30^4-sin45^4)-3(sin²60-sec²45)+1/4cot²30

(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) $\frac{2 s i n 68^{o}}{c o s 22^{o}} - \frac{2 c o t 15^{o}}{5 t a n 75^{o}} - \frac{3 t a n 45^{o} t a n 20^{o} t a n 40^{o} t a n 50^{o} t a n 70^{o}}{5}$
(viii) $\frac{3 c o s 55^{o}}{7 s i n 35^{o}} - \frac{4 (c o s 70^{o} c o s e c 20^{o})}{7 (t a n 5^{o} t a n 25^{o} t a n 45^{o} t a n 65^{o} t a n 85^{o})}$
(ix) $\frac{s i n 18^{o}}{c o s 72^{o}} + \sqrt{3} {t a n 10^{o} t a n 30^{o} t a n 40^{o} t a n 50^{o} t a n 80^{o}}$
(x) $\frac{c o s 58^{o}}{s i n 32^{o}} + \frac{s i n 22^{o}}{c o s 68^{o}} - \frac{c o s 38^{o} c o s e c 52^{o}}{t a n 18^{o} t a n 35^{o} t a n 60^{o} t a n 72^{o} t a n 55^{o}}$

Q. Find the value of x for which

|_{x}^{3}_{1}^{x} | = |_{8}^{3}_{1}^{2} |

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23cos430°-sin445°-3sin260°-sec245°+14cot230°=? (a) 5312 (b) 7324 (c) 11324 (d) 8312

(c) 11324 23cos4 30o-sin4 45o-3sin2 60o-sec2 45o+14cot2 30o=23324-124-3322-22+14×32=23916-14-334-2+34=23×516-3×-54+34=524+154+34=5+90+1824=11324

$\frac{2}{3} (\cos^{4} 30 ° - \sin^{4} 45 °) - 3 (\sin^{2} 60 ° - \sec^{2} 45 °) + \frac{1}{4} \cot^{2} 30 ° = ?$
(a) $\frac{53}{12}$
(b) $\frac{73}{24}$
(c) $\frac{113}{24}$
(d) $\frac{83}{12}$