Question

sin²π/6+cos²π/3-tan²π/4-1/2

Answer 1

hey, mate here is ur Ans
sinπ/6=1/2
cosπ/3=1/2
Tan (π/4)=1/
so,.

(1/2)²+(1/2)²-(1/)²=-(1/2)
1/4+1/4-1
1/2-1
-1/2
hence proved.

Answer 2

$\huge\bf\colorbox{aqua}{\red{Answer}}$

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

$\huge\bf\colorbox{aqua}{\red{Formula}}$

F1) sin 30° = cos 60° = $\frac {1}{2}$

F2) tan 45° = 1

$\huge\bf\colorbox{aqua}{\red{Explanation}}$

First we take LHS side

LHS:-

In trigonometry the value of π is 180°

$\frac{\pi}{6} = \frac{ \cancel{180} \: {}^{30} }{ \cancel{6}} = 30 \degree$

$\frac{\pi}{3} = \frac{ \cancel{180} \: {}^{60} }{ \cancel{3}} = 60 \degree$

$\frac{\pi}{4} = \frac{ \cancel{180} \: {}^{45} }{ \cancel{4}} = 45 \degree$

${ \sin }^{2} \frac{\pi}{6} + { \cos }^{2} \frac{\pi}{3} - {tan}^{2} \frac{\pi}{4} = { \sin }^{2} 30° + { \cos }^{2} 60° - {tan}^{2} 45°$

$= { (\frac{1}{2}) }^{2} + { (\frac{1}{2}) }^{2} - {1}^{2}$

$= \frac{1}{4} + \frac{1}{4} - 1$

$= \frac{1}{2} - 1$

$= \frac{1 - 2}{2}$

$= - \frac{1}{2} = RHS$

$\therefore{\underline{\underline{LHS=RHS}}}$

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