4- 230-1-sin260-cos 245- is

The correct option is D 13/6 4/ ^230^∘+1/sin^260^∘ cos ^245^∘ =4/(√(3))^2+1(√(3)2)^2 (1/√(2))^2 =4/3+4/3 1/2 =8/3 1/2 = 13/6 O ...

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Byju's Answer

Standard XII

Mathematics

Direction Cosines

4/ 230∘+1/sin...

Question

$\frac{4}{{cot}^{2} 30^{\circ}} + \frac{1}{{sin}^{2} 60^{\circ}} - {cos}^{2} 45^{\circ}$ is

\frac{12}{5}

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\frac{3}{4}

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\frac{13}{6}

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Solution

The correct option is D $\frac{13}{6}$
$\frac{4}{{cot}^{2} 30^{\circ}} + \frac{1}{{sin}^{2} 60^{\circ}} - {cos}^{2} 45^{\circ}$
$= \frac{4}{{(\sqrt{3})}^{2}} + \frac{1}{{(\frac{\sqrt{3}}{2})}^{2}} - {(\frac{1}{\sqrt{2}})}^{2}$
$= \frac{4}{3} + \frac{4}{3} - \frac{1}{2}$
$= \frac{8}{3} - \frac{1}{2} = \frac{13}{6}$
Option C is correct.

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Similar questions

Q. Evaluate

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

Q. Evaluate each of the following :

\frac{1}{c o t^{2} 30^{\circ}} + \frac{1}{s i n^{2} 60^{\circ}} - c o s^{2} 45^{\circ}

Verify the following :

(i) $\frac{3}{7} \times (\frac{5}{6} + \frac{12}{13}) = (\frac{3}{7} \times \frac{5}{6}) + (\frac{3}{7} \times \frac{12}{13})$
(ii) $\frac{- 15}{4} \times (\frac{3}{7} + \frac{- 12}{5}) = (\frac{- 15}{4} \times \frac{3}{7}) + (\frac{- 15}{4} \times \frac{- 12}{5})$
(iii) $(\frac{- 8}{3} + \frac{- 13}{12}) \times \frac{5}{6} = (\frac{- 8}{3} \times \frac{5}{6}) + (\frac{- 13}{12} \times \frac{5}{6})$
(iv) $\frac{- 16}{7} \times (\frac{- 8}{9} + \frac{- 7}{6}) = (\frac{- 16}{7} \times \frac{- 8}{9}) + (\frac{- 16}{7} \times \frac{- 7}{6})$

Q. Evaluate:

(i) s i n^{2} 30^{\circ} . c o s^{2} 45^{\circ} + 4 t a n^{2} 30^{\circ} + \frac{1}{2} s i n^{2} 90^{\circ} + \frac{1}{8} c o t^{2} 60^{\circ}

(i i) \frac{t a n^{2} 60^{\circ} + 4 s i n^{2} 45^{\circ} + 3 s e c^{2} 30^{\circ} + 5 c o s^{2} 90^{\circ}}{c o s e c 30^{\circ} + s e c 60^{\circ} - c o t^{2} 30^{\circ}}

[4 MARKS]

Q. Find

s i n^{2} 60^{\circ}

c o s^{2} 45^{\circ}

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4cot230∘+1sin260∘−cos245∘ is

The correct option is D 1364cot230∘+1sin260∘−cos245∘ =4(√3)2+1(√32)2−(1√2)2 =43+43−12 =83−12=136 Option C is correct.

$\frac{4}{{cot}^{2} 30^{\circ}} + \frac{1}{{sin}^{2} 60^{\circ}} - {cos}^{2} 45^{\circ}$ is