Question

Plz solve it..........✌❤

Answer 1

Before finding the required value, let us know some trigonometric values :

• tan(2nπ + θ) = tanθ, n = 1, 2, 3, ...
• tan60° = √3
• tan30° = 1/√3

Now we find the value of tan(19π/3) :

Here, 19π/3 = 6π + π/3

Taking tan on both sides, we have

tan(19π/3) = tan(6π + π/3)

= tan{3 (2π) + π/3}

= tan(π/3)

= √3

∴ tan(19π/3) = √3

Answer 2

Question:

find the value of 19π/3

Solution :

we have,

$\frac{19\pi}{3} = 6\pi + \frac{\pi}{3}$

tanx is an periodic function of 2π

⇒ it's values repeats after n2π i.e,

2π,4π ,6π .......

⇒ tan ( n2π +x ) = tan x

∴ tan (19π/3) = tan (6π +π/3)

= tan (π/3 )

= √3

which is the required solution!