Question:

\(\displaystyle \frac{\sin 30^\circ+\cos 45^\circ-\tan 60^\circ}{\cot 30^\circ-\sin 45^\circ-\cos 60^\circ}=\)

Show Hint

When many surds appear, clear small fractions (like halves) first—often reveals cancellations.
Updated On: Oct 27, 2025
  • \(-1\)
  • \(0\)
  • \(1\)
  • \(2\)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

Step 1: Substitute standard values.
\(\sin30^\circ=\tfrac12,\ \cos45^\circ=\tfrac{\sqrt2}{2},\ \tan60^\circ=\sqrt3,\ \cot30^\circ=\sqrt3,\ \cos60^\circ=\tfrac12\).
Numerator \(= \tfrac12+\tfrac{\sqrt2}{2}-\sqrt3\).
Denominator \(= \sqrt3-\tfrac{\sqrt2}{2}-\tfrac12\).
Step 2: Multiply by 2.
\(\frac{1+\sqrt2-2\sqrt3}{\,2\sqrt3-\sqrt2-1\,}=-1\) since the numerator is the negative of the denominator.
Was this answer helpful?
0
0

Questions Asked in Bihar Class X Board exam

View More Questions