0°
30°
45°
60°
sin 2A = 2 sin A.
Of the given choices, only A = 0° satisfies the equation.
For A = 0°, sin 2A = sin 0° = 0.
Also, 2 sin A = 2 sin 0° = 2(0) = 0.
Therefore, option (A) is correct.
Evaluate the following.
(i) sin60° cos30° + sin30° cos 60°
(ii) 2tan245° + cos230° - sin260°
(iii) \(\frac{cos 45°}{sec 30°+cosec30°}\)
(iv) \(\frac{sin\ 30°+tan\ 45°cosec\ 60°}{sec\ 30°+cos\ 60°+cot\ 45°}\)
(v) \(\frac{5cos^260°+4sec^230°-tan^245°}{sin^230°+cos^230°}\)