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Find the values of the trigonometric fu...

Find the values of the trigonometric function tan `(19pi)/3`

A

`sqrt3`

B

`sqrt4`

C

`sqrt6`

D

`sqrt2`

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The correct Answer is:
To find the value of the trigonometric function \( \tan\left(\frac{19\pi}{3}\right) \), we can simplify the angle using the periodic property of the tangent function. Here’s a step-by-step solution: ### Step 1: Understand the periodicity of the tangent function The tangent function has a period of \( \pi \). This means that: \[ \tan(\theta + n\pi) = \tan(\theta) \] for any integer \( n \). ### Step 2: Simplify the angle \( \frac{19\pi}{3} \) To simplify \( \frac{19\pi}{3} \), we can subtract multiples of \( 2\pi \) (which is equivalent to \( \frac{6\pi}{3} \)) to bring the angle within the range of \( 0 \) to \( 2\pi \). First, we can find how many full \( 2\pi \) (or \( \frac{6\pi}{3} \)) fit into \( \frac{19\pi}{3} \): \[ \frac{19\pi/3}{6\pi/3} = \frac{19}{6} \approx 3.1667 \] This means we can subtract \( 3 \times 2\pi \) from \( \frac{19\pi}{3} \). ### Step 3: Calculate the equivalent angle Now, we subtract \( 3 \times 2\pi \): \[ \frac{19\pi}{3} - 3 \times 2\pi = \frac{19\pi}{3} - \frac{18\pi}{3} = \frac{\pi}{3} \] ### Step 4: Find the tangent of the simplified angle Now we can find the tangent of the simplified angle: \[ \tan\left(\frac{19\pi}{3}\right) = \tan\left(\frac{\pi}{3}\right) \] ### Step 5: Use known values of tangent The value of \( \tan\left(\frac{\pi}{3}\right) \) is a known trigonometric value: \[ \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \] ### Final Answer Thus, the value of \( \tan\left(\frac{19\pi}{3}\right) \) is: \[ \boxed{\sqrt{3}} \]

To find the value of the trigonometric function \( \tan\left(\frac{19\pi}{3}\right) \), we can simplify the angle using the periodic property of the tangent function. Here’s a step-by-step solution: ### Step 1: Understand the periodicity of the tangent function The tangent function has a period of \( \pi \). This means that: \[ \tan(\theta + n\pi) = \tan(\theta) \] for any integer \( n \). ...
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NCERT ENGLISH-TRIGONOMETRIC FUNCTIONS-All Questions
  1. Prove that:(sinx-siny)/(cosx+cosy)=tan(x-y)/2

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  2. Prove that:(sinx+sin3x)/(cosx+cos3x)=tan2x

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  3. Find the values of the trigonometric function tan (19pi)/3

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  4. Find the values of the trigonometric function sin (-(11pi)/3)

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  5. Find the value of other five trigonometric function cosx=-1/2, x lies...

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  6. Find the value of other five trigonometric function sinx=3/5, x lies ...

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  7. Find the value of other five trigonometric function cotx=3/4, x lies ...

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  8. Find the value of other five trigonometric function secx=(13)/5, x li...

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  9. Find the value of other five trigonometric function tanx=-5/(12), x l...

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  10. Find the values of the trigonometric function sin 765^(@)

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  11. Find the values of the trigonometric function c o s e c(-1410^@).

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  12. cot^2pi/6+cose c(5pi)/6+3tan^2pi/6=6

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  13. 2sin^2(pi/6 )+c o s e c^2((7pi)/6)cos^2(pi/3)=3/2

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  14. sin^2pi/6+cos^2pi/3-tan^2pi/4=-1/2

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  15. Prove that:(tan(pi/4+x))/(tan(pi/4-x))=((1+tanx)/(1-tanx))^2

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  16. Prove that:cos(pi/4-x)cos(pi/4-y)-sin(pi/4-x)sin(pi/4-y)=sin(x+y)

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  17. Find the value of : (i) sin75^o (ii) tan15^o

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  18. 2sin^2(3pi)/4+2cos^2pi/4+2sec^2pi/3=10

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  19. Prove that: cos((3pi)/2+x)cos(2pi+x)[cot((3pi)/2-x)+cot(2pi+x)]=1

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  20. Prove that: (cos(pi+x)cos(-x))/(sin(pi-x)cos(pi/2+x))=cot^2x

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