Question

Prove that

Sin^2 π/6 + Cos^2 π/3 - tan^2 π/4 = - 1/2 ​

Answer 1

$\bf \: LHS = {sin}^{2} \frac{\pi}{6} + {cos}^{2} \frac{\pi}{3} - {tan}^{2} \frac{\pi}{4}$

We know that the value of π = 180°

now,

$\bf \: {sin}^{2} \frac{180}{6} + {cos}^{2} \frac{180}{3} - {tan}^{2} \frac{180}{4} \\ \\ \\ \implies \bf \: ( {sin \: 30)}^{2} + {(cos \: 60)}^{2} - {(tan \: 45)}^{2} \\ \\ \\ \implies \bf \: {( \frac{1}{2}) }^{2} + ( \frac{1}{2} ) {}^{2} - {1}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \huge \{ \tiny \: (sin30 = \frac{1}{2} ) \: \: (cos \: 60 = \frac{1}{2}) \\ \\ \\ \implies \bf \: \frac{1}{4} + \frac{1}{4} - 1 \\ \\ \\ \implies \bf \: - \frac{1}{2}$