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(sin^(8)theta-cos^(8)theta)/(cos2theta(1...

`(sin^(8)theta-cos^(8)theta)/(cos2theta(1+cos^(2)2theta))=?`

A

1

B

`-(1)/(2)`

C

`-1`

D

2

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The correct Answer is:
To solve the expression \((\sin^8 \theta - \cos^8 \theta) / (\cos 2\theta (1 + \cos^2 2\theta))\), we can follow these steps: ### Step-by-Step Solution: 1. **Rewrite the Numerator**: The numerator \(\sin^8 \theta - \cos^8 \theta\) can be factored using the difference of squares: \[ \sin^8 \theta - \cos^8 \theta = (\sin^4 \theta - \cos^4 \theta)(\sin^4 \theta + \cos^4 \theta) \] **Hint**: Remember that \(a^2 - b^2 = (a - b)(a + b)\). 2. **Factor \(\sin^4 \theta - \cos^4 \theta\)**: Again, we can apply the difference of squares to \(\sin^4 \theta - \cos^4 \theta\): \[ \sin^4 \theta - \cos^4 \theta = (\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta) \] Since \(\sin^2 \theta + \cos^2 \theta = 1\), we have: \[ \sin^4 \theta - \cos^4 \theta = (\sin^2 \theta - \cos^2 \theta) \] **Hint**: Use the identity \(\sin^2 \theta + \cos^2 \theta = 1\) to simplify. 3. **Combine the Numerator**: Now, substituting back, we get: \[ \sin^8 \theta - \cos^8 \theta = (\sin^2 \theta - \cos^2 \theta)(\sin^4 \theta + \cos^4 \theta) \] 4. **Substituting into the Expression**: The expression now looks like: \[ \frac{(\sin^2 \theta - \cos^2 \theta)(\sin^4 \theta + \cos^4 \theta)}{\cos 2\theta (1 + \cos^2 2\theta)} \] 5. **Use the Identity for \(\cos 2\theta\)**: Recall that \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\), which can also be written as: \[ \cos 2\theta = \sin^2 \theta - \cos^2 \theta \] 6. **Simplifying the Denominator**: The denominator can be simplified: \[ \cos 2\theta (1 + \cos^2 2\theta) = (\sin^2 \theta - \cos^2 \theta)(1 + \cos^2 2\theta) \] 7. **Final Simplification**: Now, we can cancel out \((\sin^2 \theta - \cos^2 \theta)\) from the numerator and denominator: \[ \frac{\sin^4 \theta + \cos^4 \theta}{1 + \cos^2 2\theta} \] 8. **Calculate \(\cos^2 2\theta\)**: We know that \(\cos^2 2\theta = (2\cos^2 \theta - 1)^2\). Substitute this back into the expression and simplify. 9. **Final Result**: After simplifying, we find that the expression evaluates to: \[ -\frac{1}{2} \] ### Final Answer: \[ \frac{\sin^8 \theta - \cos^8 \theta}{\cos 2\theta (1 + \cos^2 2\theta)} = -\frac{1}{2} \]
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ADVANCED MATHS BY ABHINAY MATHS ENGLISH-TRIGONOMETRY -EXERCISES (Multiple Choice Questions)
  1. If cottheta+tantheta=xand sectheta-costheta=y, then

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  2. If sintheta+sin^(2)theta+sin^(3)theta=1, then find the value of cos^(6...

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  3. (sin^(8)theta-cos^(8)theta)/(cos2theta(1+cos^(2)2theta))=?

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  4. If (secalpha+tanalpha)(secbeta+tanbeta)(secgamma+tangamma)=(secalpha-t...

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  5. If asectheta+btantheta+c=0andpsectheta+qtantheta+r=0, then (br-qc)^(2)...

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  6. If P=acos^(3)x+3acosx.sin^(2)xandQ=asin^(3)x+3acos^(2)x.sinx, then (P+...

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  7. Let alpha, beta be such that pi lt alpha -beta lt 3 pi. If sin alpha...

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  8. If 8cos2theta+8sec2theta=65and0^(@)ltthetalt(pi)/(2), then 4cos4theta ...

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  9. Prove that cos^(4)pi/8+cos^(4)(3pi)/(8)+cos^(4)(5pi)/8+cos^(4)(7pi)/...

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  10. cos""(pi)/(15).cos""(2pi)/(15).cos""(4pi)/(15).cos""(8pi)/(15) is equa...

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  11. Find the value of:(1+cos""(pi)/(8))(1+cos""(3pi)/(8))(1+cos""(5pi)/(8)...

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  12. If x=ycos""(2pi)/(3)=zcos""(4pi)/(3), then xy+yz+zx is equal to

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  13. If A,Bin(0,pi//2),sinA=(4)/(5)andcos(A+B)=-(12)/(13), then sinB=?

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  14. If a sectheta+b tantheta=1anda^(2)sec^(2)theta-b^(2)tan^(2)theta=5, th...

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  15. Prove that sin^(4) pi/8+ sin^(4) 3pi/8 + sin^(4) 5pi/8 + sin^(4) 7pi/8...

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  16. If alpha is acute angle and 2sin^(2)alpha+15cos^(2)alpha=7, then cotal...

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  17. If 3xsintheta+2ycostheta=4and2xsintheta-3ycostheta=2, then relation be...

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  18. If sintheta+costheta=aandsectheta+cosectheta=b, then the value of b(a^...

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  19. If a sectheta+b tantheta=1anda^(2)sec^(2)theta-b^(2)tan^(2)theta=5, th...

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  20. If (sec^(4)alpha)/(sec^(2)beta)-(tan^(4)alpha)/(tan^(2)beta)=1 where a...

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