$The value of Sin 30 ° + \cos 60 ° - \tan 45 °$

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$0$

$\frac{1}{2}$

$\sqrt{\frac{3}{2}}$

answer is D.

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Detailed Solution

$The value of the given trigonometric functions \sin 30 ° + \cos 60 ° - \tan 45 ° is zero$

$G i v e n, \sin 30 ° + \cos 60 ° - \tan 45 °$

$T o f i n d, t h e v a l u e o f t h e g i v e n t r i g o n o m e t r i c e x p r e s s i o n$ $T h e r e s u l \tan t v a l u e o f t h e s u m o f a l l t h e s e t r i g o n o m e t r i c f u n c t i o n s i s g i v e n t o u s a s a t r i g o n o m e t r i c e x p r e s s i o n w i t h a n g l e s m e a s u r e d i n d e g r e e u n i t s . L e t' s s t a r t b y d e t e r m i n i n g t h e k n o w n v a l u e s o f e a c h o f t h e s e t r i g o n o m e t r i c f u n c t i o n s u \sin g t h e a n g l e s s u p p l i e d .$ $\sin 30 ° = 1 / 2$

$\cos 60 ° = 1 / 2$

$\tan 45 ° = 1$

$s u b s t i t u t i n g t h e s e v a l u e i n t h e g i v e n q u e s t i o n, t h a t i s,$

$\sin 30 ° + \cos 60 ° - \tan 45 °$
$= 1 / 2 + 1 / 2 - 1$

$= {(1 + 1) / 2} - 1$

$= 1 - 1$

$= 0$

$T h e r e f o r e, t h e t r i g o n o m e t r i c f u n c t i o n s \sin 30 ° + \cos 60 ° - \tan 45 ° h a v e z e r o v a l u e a s a r e s u l t .$