Question

prove that
sin2,
$\sin(2) \binom{\pi}{6} + \cos(2) - \tan(2) \binom{\pi}{4} = - \binom{1}{2}$
​

Answer 1

⇝Correct question:-

$\sin(2) \binom{\pi}{6} + \cos(2 \binom{\pi}{3} ) - \tan(2) \binom{\pi}{4 } = - \binom{1}{2}$

⇝Answer:-

• By solving L.H.S

$\sin(2) \binom{\pi}{6} + \cos(2 \binom{\pi}{3} ) - \tan(2) \binom{\pi}{4 }$

• Putting
• $\pi = 180$

$\sin(2) \binom{180}{6} + \cos(2) \binom{180}{3} - \tan(2) \binom{180}{4}$

$\sin(2) 30 + \cos(2) 60 - \tan(2) 45$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━

⇝Now by adding :-

• sin 30= 1/2
• cos 60= 1/2
• tan 45= 1

━━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\binom{1}{2} {}^{2} + \binom{1}{2} {}^{2} - 1 {}^{2}$

$\binom{1}{4} + \binom{1}{4} - 1$

$\binom{1 + 1 - 4}{4}$

$\binom{2 - 4}{4}$

$\binom{ - 2}{4}$

$\binom{ - 1}{2}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━

⇝R.H.S :-

$\sin(2) \binom{\pi}{6} + \cos(2 \binom{\pi}{3} ) - \tan(2) \binom{\pi}{4 } = - \binom{1}{2}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━

⇝Proved !