The value of sin 2-6-cos 2-3-tan 2-4- 2-2 is equal

The correct option is A 12sin^2π6+cos^2π3 tan^2π4+^2π2 =(12)^2+(12)^2 (1)^2+0 = 12 ...

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Sign of Trigonometric Ratios in Different Quadrants

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Question

The value of ${sin}^{2} \frac{π}{6} + {cos}^{2} \frac{π}{3} - {tan}^{2} \frac{π}{4} + {cot}^{2} \frac{π}{2}$ is equal to

- \frac{1}{2}

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1

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\frac{\sqrt{3}}{2}

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0

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{sin}^{2} \frac{π}{6} + {cos}^{2} \frac{π}{3} - {tan}^{2} \frac{π}{4} = - \frac{1}{2}

{sin}^{2} \frac{π}{6} + {cos}^{2} \frac{π}{3} - {tan}^{2} \frac{π}{4} = - \frac{1}{2}

sin \frac{π}{6} cos 0 + sin \frac{π}{4} cos \frac{π}{4} + sin \frac{π}{3} cos \frac{π}{6} = \frac{7}{4}

{sin}^{2} \frac{π}{6} + {cos}^{2} \frac{π}{3} - {tan}^{2} \frac{π}{4} = - \frac{1}{2}

4 {cos}^{2} \frac{π}{3} + {sec}^{2} \frac{π}{6} - {sin}^{2} \frac{π}{4}

{⎡ ⎣ \begin{matrix} cos \frac{2 π}{3} & - sin \frac{2 π}{3} sin \frac{2 π}{3} & cos \frac{2 π}{3} \end{matrix} ⎤ ⎦}^{k} = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]

k

(k \neq 0)

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