The value of 4 cot2-3-sec2-6-sin2-4 is

The correct option is B 136 4 cot^2π/3+sec^2π/6 sin^2π/4=4( 1 /√( 3 ) nbsp; nbsp;) nbsp;^ 2 +( 2 /√( 3 ) nbsp; nbsp;) nbsp;^ 2 ( 1 ...

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Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

The value of ...

Question

The value of 4 $c o t^{2} \frac{π}{3} + s e c^{2} \frac{π}{6} - s i n^{2} \frac{π}{4}$ is

\frac{13}{6}

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\frac{11}{6}

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\frac{7}{6}

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\frac{5}{6}

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Solution

The correct option is B $\frac{13}{6}$
4 $c o t^{2} \frac{π}{3} + s e c^{2} \frac{π}{6} - s i n^{2} \frac{π}{4} = 4 {(\frac{1}{\sqrt{3}})}^{2} + {(\frac{2}{\sqrt{3}})}^{2} - {(\frac{1}{\sqrt{2}})}^{2} = \frac{4}{3} + \frac{4}{3} - \frac{1}{2} = \frac{16 - 3}{6} = \frac{13}{6}$

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Similar questions

Q. Question 6
Explain why

7 \times 11 \times 13 + 13

and

7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5

are composite numbers.

Q. Calculate the values of the determinants:

∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 7 & 13 & 10 & 6 5 & 9 & 7 & 4 8 & 12 & 11 & 7 4 & 10 & 6 & 3 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

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. Find

(\frac{3}{11} \times \frac{5}{6}) - (\frac{9}{12} \times \frac{4}{3}) + (\frac{5}{13} \times \frac{6}{15})

\frac{(- \frac{5}{6})^{3 / 4}}{(- \frac{5}{6})^{7 / 6}} = (- \frac{5}{6})^{7 - x}

, the value of x is :

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The value of 4 cot2π3+sec2π6−sin2π4 is

The correct option is B 1364 cot2π3+sec2π6−sin2π4=4(1√3)2+(2√3)2−(1√2)2=43+43−12=16−36=136

The value of 4 $c o t^{2} \frac{π}{3} + s e c^{2} \frac{π}{6} - s i n^{2} \frac{π}{4}$ is

The correct option is B $\frac{13}{6}$
4 $c o t^{2} \frac{π}{3} + s e c^{2} \frac{π}{6} - s i n^{2} \frac{π}{4} = 4 {(\frac{1}{\sqrt{3}})}^{2} + {(\frac{2}{\sqrt{3}})}^{2} - {(\frac{1}{\sqrt{2}})}^{2} = \frac{4}{3} + \frac{4}{3} - \frac{1}{2} = \frac{16 - 3}{6} = \frac{13}{6}$