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The vlaue of cosec^(2)""(pi)/(7)+cosec^(...

The vlaue of `cosec^(2)""(pi)/(7)+cosec^(2)""(2pi)/(7)+cosec^(2)""(3pi)/(7),` is

A

20

B

2

C

22

D

23

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The correct Answer is:
To find the value of \( \csc^2\left(\frac{\pi}{7}\right) + \csc^2\left(\frac{2\pi}{7}\right) + \csc^2\left(\frac{3\pi}{7}\right) \), we can follow these steps: ### Step 1: Rewrite Cosecant in Terms of Sine We know that: \[ \csc^2 \theta = \frac{1}{\sin^2 \theta} \] Thus, we can rewrite the expression as: \[ \csc^2\left(\frac{\pi}{7}\right) + \csc^2\left(\frac{2\pi}{7}\right) + \csc^2\left(\frac{3\pi}{7}\right) = \frac{1}{\sin^2\left(\frac{\pi}{7}\right)} + \frac{1}{\sin^2\left(\frac{2\pi}{7}\right)} + \frac{1}{\sin^2\left(\frac{3\pi}{7}\right)} \] ### Step 2: Use the Identity for Sine We can use the identity: \[ \sin^2 x = \frac{1 - \cos(2x)}{2} \] So, we can rewrite each sine term: \[ \sin^2\left(\frac{\pi}{7}\right) = \frac{1 - \cos\left(\frac{2\pi}{7}\right)}{2}, \quad \sin^2\left(\frac{2\pi}{7}\right) = \frac{1 - \cos\left(\frac{4\pi}{7}\right)}{2}, \quad \sin^2\left(\frac{3\pi}{7}\right) = \frac{1 - \cos\left(\frac{6\pi}{7}\right)}{2} \] ### Step 3: Substitute Back into the Expression Now substituting back, we have: \[ \frac{1}{\sin^2\left(\frac{\pi}{7}\right)} = \frac{2}{1 - \cos\left(\frac{2\pi}{7}\right)}, \quad \frac{1}{\sin^2\left(\frac{2\pi}{7}\right)} = \frac{2}{1 - \cos\left(\frac{4\pi}{7}\right)}, \quad \frac{1}{\sin^2\left(\frac{3\pi}{7}\right)} = \frac{2}{1 - \cos\left(\frac{6\pi}{7}\right)} \] ### Step 4: Combine the Terms Thus, the expression becomes: \[ \csc^2\left(\frac{\pi}{7}\right) + \csc^2\left(\frac{2\pi}{7}\right) + \csc^2\left(\frac{3\pi}{7}\right) = 2 \left( \frac{1}{1 - \cos\left(\frac{2\pi}{7}\right)} + \frac{1}{1 - \cos\left(\frac{4\pi}{7}\right)} + \frac{1}{1 - \cos\left(\frac{6\pi}{7}\right)} \right) \] ### Step 5: Use the Sum of Cosines Using the identity for the sum of cosines, we can simplify the expression further. The sum of cosines can be expressed as: \[ \cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right) = -\frac{1}{2} \] ### Step 6: Final Calculation Now we can substitute these values into the expression and simplify to find the final result. After performing the calculations, we find that: \[ \csc^2\left(\frac{\pi}{7}\right) + \csc^2\left(\frac{2\pi}{7}\right) + \csc^2\left(\frac{3\pi}{7}\right) = 8 \] Thus, the final answer is: \[ \boxed{8} \]
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OBJECTIVE RD SHARMA-TRIGONOMETRIC RATIOS AND IDENTITIES-Exercise
  1. The value of cos""(pi)/(9)cos""(2pi)/(9)cos""(3pi)/(9), is

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  2. The value of cos""(pi)/(9)cos""(2pi)/(9)cos""(3pi)/(9)cos""(4pi)/(9), ...

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  3. The vlaue of cosec^(2)""(pi)/(7)+cosec^(2)""(2pi)/(7)+cosec^(2)""(3pi)...

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  4. sin1 2^(@)sin4 8^(@)sin5 4^(@)=

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  5. the value of sin(pi/7)+sin((2pi)/7)+sin((3pi)/7) is

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  6. cos^6pi/9-33tan^4pi/9+27tan^2pi/9 is equal to 0 (b) sqrt(3) (c) 3 ...

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  7. (sin^2 3A)/(sin^2A)-(cos^2 3A)/(cos^2A)=

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  8. If sinA=(336)/(625)where450^(@)ltAlt540^(@), then sin""(A)/(4)=

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  9. If y=(tanx)/(tan3x), then

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  10. The value of cot^(2)""(pi)/(7)+cot^(2)""(2pi)/(7)+cot^(2)""(3pi)/(7), ...

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  11. The value of sin""(pi)/(7)sin""(2pi)/(7)sin""(3pi)/(7), is

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  12. The value of sin""(2pi)/(7)+sin""(4pi)/(7)+sin""(8pi)/(7), is

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  13. cos""(2pi)/(15)cos""(4pi)/(15)cos""(8pi)/(15)cos""(16pi)/(15)=(1)/(16)

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  14. If sinA+cosA=m an sin^(3)A+cos^(3)A=n, then

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  15. If cosA+cosB=m and sinA+sinB=n then sin(A+B)=

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  16. If ltA lt(pi)/(6) and sinA+cosA=(sqrt7)/(2),"then" tan""(A)/(2)=

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  17. The value of cos""(pi)/(11)+cos""(3pi)/(11)+cos""(5pi)/(11)+cos""(7pi)...

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  18. If n=pi/(4alpha), then tanalpha tan 2alpha tan 3 alpha........tan(2n-1...

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  19. the value of tan9^@-tan2 7^@-tan6 3^@+tan8 1^@ is equal to

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  20. For x in R, tanx+1/2tan""(1)/(2^(2))tan""(x)/(2^(2))+...+(1)/(2^(n-1...

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