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(kcosec^(2)30^(@).sec^(2)45)/(8cos^(2)45...

`(kcosec^(2)30^(@).sec^(2)45)/(8cos^(2)45^(@).sin^(2)60^(@))=tan^(2)60^(@)-tan^(2)30^(@)`, then k = ?

A

1

B

`-1`

C

2

D

0

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The correct Answer is:
To solve the equation \[ \frac{k \cdot \cos^2(30^\circ) \cdot \sec^2(45^\circ)}{8 \cdot \cos^2(45^\circ) \cdot \sin^2(60^\circ)} = \tan^2(60^\circ) - \tan^2(30^\circ) \] we will start by substituting the known values of the trigonometric functions. ### Step 1: Substitute the known values 1. **Values of trigonometric functions**: - \(\cos(30^\circ) = \frac{\sqrt{3}}{2} \Rightarrow \cos^2(30^\circ) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}\) - \(\sec(45^\circ) = \sqrt{2} \Rightarrow \sec^2(45^\circ) = (\sqrt{2})^2 = 2\) - \(\cos(45^\circ) = \frac{1}{\sqrt{2}} \Rightarrow \cos^2(45^\circ) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}\) - \(\sin(60^\circ) = \frac{\sqrt{3}}{2} \Rightarrow \sin^2(60^\circ) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}\) - \(\tan(60^\circ) = \sqrt{3} \Rightarrow \tan^2(60^\circ) = (\sqrt{3})^2 = 3\) - \(\tan(30^\circ) = \frac{1}{\sqrt{3}} \Rightarrow \tan^2(30^\circ) = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}\) Now substituting these values into the equation: \[ \frac{k \cdot \frac{3}{4} \cdot 2}{8 \cdot \frac{1}{2} \cdot \frac{3}{4}} = 3 - \frac{1}{3} \] ### Step 2: Simplify the left side 2. **Simplifying the left side**: \[ \frac{k \cdot \frac{3}{4} \cdot 2}{8 \cdot \frac{1}{2} \cdot \frac{3}{4}} = \frac{k \cdot \frac{3}{2}}{8 \cdot \frac{1}{2} \cdot \frac{3}{4}} \] Calculating the denominator: \[ 8 \cdot \frac{1}{2} \cdot \frac{3}{4} = 8 \cdot \frac{3}{8} = 3 \] So the left side becomes: \[ \frac{k \cdot \frac{3}{2}}{3} = \frac{k}{2} \] ### Step 3: Simplify the right side 3. **Simplifying the right side**: Calculating \(3 - \frac{1}{3}\): \[ 3 - \frac{1}{3} = \frac{9}{3} - \frac{1}{3} = \frac{8}{3} \] ### Step 4: Set the simplified expressions equal 4. **Setting the two sides equal**: Now we have: \[ \frac{k}{2} = \frac{8}{3} \] ### Step 5: Solve for \(k\) 5. **Solving for \(k\)**: Multiply both sides by 2: \[ k = \frac{8}{3} \cdot 2 = \frac{16}{3} \] ### Final Answer Thus, the value of \(k\) is: \[ k = \frac{16}{3} \]
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ADVANCED MATHS BY ABHINAY MATHS ENGLISH-TRIGONOMETRY -EXERCISES (Multiple Choice Questions)
  1. (cos(90^(@)-theta).sec(90^(@)-theta).tantheta)/(cosec(90^(@)-theta).si...

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  2. If alpha is in first quadrant such that tan^(2)alpha=(8)/(7), then the...

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  3. (kcosec^(2)30^(@).sec^(2)45)/(8cos^(2)45^(@).sin^(2)60^(@))=tan^(2)60^...

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  4. If alpha is in Ist quadrant such that sec^(2)alpha=3, then (tan^(2)alp...

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  5. If 5alpha and 4alpha are in Ist quadrant such that sin5alpha=cos4alpha...

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  6. If (sintheta+cosectheta)^(2)+(costheta+sectheta)^(2)=K+tan^(2)theta+co...

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  7. (sinalpha+secalpha)^(2)+(cosalpha+cosecalpha)^(2)=(K+secalphacosecalph...

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  8. (sintheta)/(cottheta+cosectheta)-(sintheta)/(cottheta-cosectheta)=?

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  9. If (tanA)/(1-cotA)+(cotA)/(1-tanA)=K+tanA+cotA then K = ?

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  10. If (cos^(2)theta)/(1-tantheta)+(sin^(3)theta)/(sintheta-costheta)=K+si...

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  11. ((1-sintheta+costheta)^(2))/((1+costheta)(1-sintheta))=?

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  12. sec^(6)theta-tan^(6)theta-3tan^(2)theta.sec^(2)theta=?

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  13. cosec^(6)theta-cot^(6)theta-3cot^(2)theta.cosec^(2)theta=?

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  14. ((cosectheta-sectheta)(cottheta-tantheta))/((cosectheta+sectheta)(sect...

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  15. sec^(4)alpha(1-sin^(4)alpha)-2tan^(2)alpha=?

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  16. If sintheta+costheta=sqrt(2)sin(90^(@)-theta), then cottheta=?

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  17. If cotalpha=(15)/(8), then ((2+2sinalpha)(1-sinalpha))/((1+cosalpha)(2...

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  18. If x=asinalphaandy=bcosalpha, then b^(2)x^(2)+a^(2)y^(2)=?

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  19. ((cottheta)/(cottheta-cot3theta)+(tantheta)/(tantheta-tan3theta))=?

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  20. If tantheta-cottheta=(119)/(60)" for "0^(@)ltthetaltpi//2, then the va...

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