Question

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$\sf{\large{Prove:}}$
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$\sin ^{2} \frac{\pi }{6} + \cos ^{2} \frac{\pi }{3} – \tan ^{2} \frac{\pi }{4} = -\frac{1}{2}$

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Answer 1

sin pie/6 = 1/2

cos pie/3 = 1/2

tan pie/4 = 1

So sin^2 pie/6 + cos^2 pie/3 - tan^2 pie/4

1/4 + 1/4 - 1

1/2 - 1

-1/2

Answer 2

Answer: PROVING

Step-by-step explanation:

L. H. S. = Sin ^2 π/6 + cos^2 π/3 - tan^2 π/4

=> ( sin 30 °) ^2 + ( cos 60° ) ^2 - ( tan 45° ) ^2

=> ( 1 / 2 ) ^2 + ( 1 /2 ) ^2 - 1

=> ( 1 /4 ) + ( 1 / 4) - 1

=> ( 2 /4 ) - 1

=> ( 1 /2 ) - 1

=> ( 1 - 2 ) /2

=> - 1 /2

=> R. H. S.

HENCE, PROVED.