0°
30°
45°
60°
sin 2A = 2 sin A
Out of the given alternatives, only A = 0° is correct.
As, sin 2A = sin 0° = 0
2 sin A = 2sin 0° = 2(0) = 0
Hence, (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) \(\frac{5cos^260°+4sec^230°-tan^245°}{sin^230°+cos^230°}\)
| Case No. | Lens | Focal Length | Object Distance |
|---|---|---|---|
| 1 | \(A\) | 50 cm | 25 cm |
| 2 | B | 20 cm | 60 cm |
| 3 | C | 15 cm | 30 cm |
The relationship between the sides and angles of a right-angle triangle is described by trigonometry functions, sometimes known as circular functions. These trigonometric functions derive the relationship between the angles and sides of a triangle. In trigonometry, there are three primary functions of sine (sin), cosine (cos), tangent (tan). The other three main functions can be derived from the primary functions as cotangent (cot), secant (sec), and cosecant (cosec).
sin x = a/h
cos x = b/h
tan x = a/b
Tan x can also be represented as sin x/cos x
sec x = 1/cosx = h/b
cosec x = 1/sinx = h/a
cot x = 1/tan x = b/a
