Question

$2\sin 30^o +2\tan 45^o -3\cos 60^o -\cos^2 30^o=$
A. 0
B. 1
C. -1
D. None of these

Answer 1

Hint: Here we use the basic trigonometric ratios to find the required value. Remember basic trigonometric values.
$\sin 30^o =\dfrac{1}{2}, \tan 45^o =1, \cos 60^o =\dfrac{1}{2}, \cos 30^o =\dfrac{\sqrt3}{2}$

Complete step-by-step answer:
We have to find the value of $2\sin 30^o +2\tan 45^o -3\cos 60^o -\cos^2 30^o$.
As we know, $\sin 30^o =\dfrac{1}{2}, \tan 45^o =1, \cos 60^o =\dfrac{1}{2}, \cos 30^o =\dfrac{\sqrt3}{2}$, thus, putting these values in the above equation, we get,
$2\sin 30^o +2\tan 45^o -3\cos 60^o -\cos^2 30^o= 2\times \dfrac{1}{2}+2\times 1 -3\times \dfrac{1}{2}-\left(\dfrac{\sqrt3}{2}\right)^2$
$\implies 2\sin 30^o +2\tan 45^o -3\cos 60^o -\cos^2 30^o=1+2-\dfrac{3}{2}-\dfrac{3}{4}$
$\implies 2\sin 30^o +2\tan 45^o -3\cos 60^o -\cos^2 30^o =3-\dfrac{3}{2}-\dfrac{3}{4}$
Taking LCM in the denominator to simplify, we get,
$2\sin 30^o +2\tan 45^o – 3\cos 60^o -\cos^2 30^o = \dfrac{3\times 4-3\times 2 -3}{4}$
$\implies 2 \sin 30^o +2\tan 45^o -3\cos 60^o -\cos^2 30^o =\dfrac{12-6-3}{4}=\dfrac{3}{4}$
Thus, $2\sin 30^o +2\tan 45^o -3\cos 60^o -\cos^2 30^o = \dfrac{3}{4}$
Hence, option D is correct.

Note: In this type of questions, we just need to put the values of the basic trigonometric ratios being asked and then simplify the equation in order to find the required value.