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sin^(2) (pi/6)+cos^(2) (pi/3)-tan^(2) (p...

`sin^(2) (pi/6)+cos^(2) (pi/3)-tan^(2) (pi/4)=-1/2`

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To solve the equation \( \sin^2\left(\frac{\pi}{6}\right) + \cos^2\left(\frac{\pi}{3}\right) - \tan^2\left(\frac{\pi}{4}\right) = -\frac{1}{2} \), we will evaluate each trigonometric function step by step. ### Step 1: Evaluate \( \sin^2\left(\frac{\pi}{6}\right) \) We know that: \[ \sin\left(\frac{\pi}{6}\right) = \sin(30^\circ) = \frac{1}{2} \] ...
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Knowledge Check

  • The value of 2"sin"^(2)(pi)/(6)+"cosec"^(2)(7pi)/(6)."cos"^(2)(pi)/(3) is equal to

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    B
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