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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...

2sin⁡30o+2tan⁡45o−3cos⁡60o−cos⁡230o=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

Explanation

Solution

Here we use the basic trigonometric ratios to find the required value. Remember basic trigonometric values.
sin⁡30o=12,tan⁡45o=1,cos⁡60o=12,cos⁡30o=32\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 2sin⁡30o+2tan⁡45o−3cos⁡60o−cos⁡230o2\sin 30^o +2\tan 45^o -3\cos 60^o -\cos^2 30^o.
As we know, sin⁡30o=12,tan⁡45o=1,cos⁡60o=12,cos⁡30o=32\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,
2sin⁡30o+2tan⁡45o−3cos⁡60o−cos⁡230o=2×12+2×1−3×12−(32)22\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
  ⟹  2sin⁡30o+2tan⁡45o−3cos⁡60o−cos⁡230o=1+2−32−34\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}
  ⟹  2sin⁡30o+2tan⁡45o−3cos⁡60o−cos⁡230o=3−32−34\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,
2sin⁡30o+2tan⁡45o–3cos⁡60o−cos⁡230o=3×4−3×2−342\sin 30^o +2\tan 45^o – 3\cos 60^o -\cos^2 30^o = \dfrac{3\times 4-3\times 2 -3}{4}
  ⟹  2sin⁡30o+2tan⁡45o−3cos⁡60o−cos⁡230o=12−6−34=34\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, 2sin⁡30o+2tan⁡45o−3cos⁡60o−cos⁡230o=342\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.