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Mathematics Question on Trigonometric Functions of Sum and Difference of Two Angles

Prove that. sin2π6+cos2π3–tan2π4=−12sin^2 \frac{π}{6}+cos^2 \frac{π}{3}–tan^2 \frac{π}{4}=-\frac{1}{2}

Answer

L.H.S. = sin2π6+cos2π3–tan2π4sin^2 \frac{π}{6}+cos^2 \frac{π}{3}–tan^2 \frac{π}{4}

(12)2+(12)2−(1)2(\frac{1}{2})^2+(\frac{1}{2})^2-(1)^2

14+14−1=−12\frac{1}{4}+\frac{1}{4}-1=-\frac{1}{2}

=R.H.S=R.H.S