Prove that 

Q(1) \small \sin ^{2} \left ( \frac{\pi }{6} \right ) + \cos ^{2}\left ( \frac{\pi }{3} \right ) - \tan ^{2}\left ( \frac{\pi }{4} \right ) = -\frac{1}{2}

Answers (1)

We know the values of sin (30 degree), cos (60 degree) and tan (45 degree). That is: 

   
\sin \left ( \frac{\pi}{6} \right ) = \left ( \frac{1}{2} \right )\\ \\ \cos \left ( \frac{\pi}{3} \right ) = \left ( \frac{1}{2} \right )\\ \\ \tan \left ( \frac{\pi}{4} \right ) = 1
\sin^{2}\frac{\pi}{6}+\cos^{2}\frac{\pi}{3}-\tan^{2}\frac{\pi}{4}=   \left ( \frac{1}{2} \right )^{2}+ \left ( \frac {1}{2} \right )^{2}-1^{2}
                                                          
                                                       = \frac{1}{4}+\frac{1}{4}-1= -\frac{1}{2}
                                                       = R.H.S.

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