3 cos 2 60-2 2 30-5 sin 2 45-a 136b 174cd 4

(b) 174 3 #160;cos2 #160;60o+2 #160;cot2 #160;30o 5 #160;sin2 #160;45o #160;= #160;3 #215;122+2 #215;32 5 #215;122 #160;= #160;34+6 5 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Differentiability

3 cos 2 60∘+2...

Question

(3cos²60° + 2cot²30° − 5sin²45°) = ?
(a) $\frac{13}{6}$
(b) $\frac{17}{4}$
(c) 1
(d) 4

Open in App

Solution

(b) $\frac{17}{4}$

$(3 \cos^{2} 60^{o} + 2 \cot^{2} 30^{o} - 5 \sin^{2} 45^{o}) = [3 \times {(\frac{1}{2})}^{2} + 2 \times {(\sqrt{3})}^{2} - 5 \times {(\frac{1}{\sqrt{2}})}^{2}] = [\frac{3}{4} + 6 - \frac{5}{2}] = \frac{3 + 24 - 10}{4} = \frac{17}{4}$

Suggest Corrections

Similar questions

$3 c o s^{2} 60^{o} + 2 c o t^{2} 30^{o} - 5 s i n^{2} 45^{o}$ = ?

(a) $\frac{13}{6}$ (a) $\frac{17}{4}$ (c) 1 (d) 4

Q. (3cos²60° + 2cot²30° – 5sin²45°) = ?
(a) 1

(b) 4

(c)

\frac{17}{4}

(d)

\frac{13}{6}

Q. Find he value of:

3 {cosec}^{2} 60^{\circ} - 2 {cot}^{2} 30^{\circ} + {sec}^{2} 45^{\circ}

The value of
$\frac{c o t^{2} 30^{\circ} + 8 s i n^{2} 45^{\circ} + \frac{3}{2} s e c^{2} 30^{\circ} + 2 c o s^{2} 90^{\circ}}{2 s e c 60^{\circ} + 3 c o s e c 30^{\circ} - \frac{7}{3} t a n^{2} 60^{\circ}}$ is :

Find the value of $\frac{4}{3} t a n^{2} 60^{\circ} + 3 c o s^{2} 30^{\circ} - 2 s e c^{2} 30^{\circ} - \frac{3}{4} c o t^{2} 60^{\circ}$

Join BYJU'S Learning Program

(3cos260° + 2cot230° − 5sin245°) = ? (a) 136 (b) 174 (c) 1 (d) 4

(b) 174 3 cos2 60o+2 cot2 30o-5 sin2 45o = 3×122+2×32-5×122 = 34+6-52=3+24-104=174

(3cos²60° + 2cot²30° − 5sin²45°) = ?
(a) $\frac{13}{6}$
(b) $\frac{17}{4}$
(c) 1
(d) 4

(b) $\frac{17}{4}$

$(3 \cos^{2} 60^{o} + 2 \cot^{2} 30^{o} - 5 \sin^{2} 45^{o}) = [3 \times {(\frac{1}{2})}^{2} + 2 \times {(\sqrt{3})}^{2} - 5 \times {(\frac{1}{\sqrt{2}})}^{2}] = [\frac{3}{4} + 6 - \frac{5}{2}] = \frac{3 + 24 - 10}{4} = \frac{17}{4}$