Question

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

Answer 1

we know,
$\bf{sin\frac{\pi}{6}=\frac{1}{2}}\\\\\bf{cos\frac{\pi}{3}=\frac{1}{2}}\\\\\bf{tan\frac{\pi}{4}=1}$

now,
LHS = sin² π/6 + cos² π/3 - tan² π/4

= (1/2)² + (1/2)² - 1²

= 1/4 + 1/4 - 1

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

= 1/2 - 1

= -1/2 = RHS

hence proved