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Trigonometric Functions and Equations question

2025 · Shift 2 · Q31
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  5. /2025 · Shift 2 · Q31

Trigonometric Functions and Equations question

2025 · Shift 2 · Q31

JEE AdvancedMathematicsTrigonometric Functions and EquationsNumerical+4 / −1
Let α=1sin⁡60∘sin⁡61∘+1sin⁡62∘sin⁡63∘+⋯+1sin⁡118∘sin⁡119∘\alpha=\frac{1}{\sin 60^{\circ} \sin 61^{\circ}}+\frac{1}{\sin 62^{\circ} \sin 63^{\circ}}+\cdots+\frac{1}{\sin 118^{\circ} \sin 119^{\circ}}α=sin60∘sin61∘1​+sin62∘sin63∘1​+⋯+sin118∘sin119∘1​ Then the value of (cosec⁡1∘α)2\left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}\right)^2(αcosec1∘​)2 is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 3

Step-by-step Derivation:

  1. Analyze the expression for α: The given expression is: α=1sin⁡60∘sin⁡61∘+1sin⁡62∘sin⁡63∘+⋯+1sin⁡118∘sin⁡119∘\alpha=\frac{1}{\sin 60^{\circ} \sin 61^{\circ}}+\frac{1}{\sin 62^{\circ} \sin 63^{\circ}}+\cdots+\frac{1}{\sin 118^{\circ} \sin 119^{\circ}}α=sin60∘sin61∘1​+sin62∘sin63∘1​+⋯+sin118∘sin119∘1​ This is a sum which can be written in summation notation. The arguments of the sine functions are of the form (60+2k) and (61+2k). The first term corresponds to k=0 and the last term (118, 119) corresponds to 60+2k = 118, which gives 2k=58, so k=29. Thus, the sum is: α=∑k=0291sin⁡(60+2k)∘sin⁡(61+2k)∘\alpha = \sum_{k=0}^{29} \frac{1}{\sin(60+2k)^{\circ} \sin(61+2k)^{\circ}}α=∑k=029​sin(60+2k)∘sin(61+2k)∘1​

  2. Simplify the general term: We use the identity derived from the compound angle formula for sine: sin⁡(B−A)=sin⁡Bcos⁡A−cos⁡Bsin⁡A\sin(B-A) = \sin B \cos A - \cos B \sin Asin(B−A)=sinBcosA−cosBsinA Dividing by sin⁡Asin⁡B\sin A \sin BsinAsinB, we get: sin⁡(B−A)sin⁡Asin⁡B=sin⁡Bcos⁡Asin⁡Asin⁡B−cos⁡Bsin⁡Asin⁡Asin⁡B=cot⁡A−cot⁡B\frac{\sin(B-A)}{\sin A \sin B} = \frac{\sin B \cos A}{\sin A \sin B} - \frac{\cos B \sin A}{\sin A \sin B} = \cot A - \cot BsinAsinBsin(B−A)​=sinAsinBsinBcosA​−sinAsinBcosBsinA​=cotA−cotB Rearranging this, we get a useful formula to split the term: 1sin⁡Asin⁡B=cot⁡A−cot⁡Bsin⁡(B−A)\frac{1}{\sin A \sin B} = \frac{\cot A - \cot B}{\sin(B-A)}sinAsinB1​=sin(B−A)cotA−cotB​

  3. Apply the identity to the sum: For each term in the sum for α, we have A=(60+2k)∘A = (60+2k)^{\circ}A=(60+2k)∘ and B=(61+2k)∘B = (61+2k)^{\circ}B=(61+2k)∘. The difference B−A=1∘B - A = 1^{\circ}B−A=1∘, which is constant for all terms. Therefore, sin⁡(B−A)=sin⁡1∘\sin(B-A) = \sin 1^{\circ}sin(B−A)=sin1∘. Applying the identity, each term becomes: 1sin⁡(60+2k)∘sin⁡(61+2k)∘=cot⁡(60+2k)∘−cot⁡(61+2k)∘sin⁡1∘\frac{1}{\sin(60+2k)^{\circ} \sin(61+2k)^{\circ}} = \frac{\cot(60+2k)^{\circ} - \cot(61+2k)^{\circ}}{\sin 1^{\circ}}sin(60+2k)∘sin(61+2k)∘1​=sin1∘cot(60+2k)∘−cot(61+2k)∘​ Now, we can rewrite α as: α=1sin⁡1∘∑k=029[cot⁡(60+2k)∘−cot⁡(61+2k)∘]\alpha = \frac{1}{\sin 1^{\circ}} \sum_{k=0}^{29} \left[ \cot(60+2k)^{\circ} - \cot(61+2k)^{\circ} \right]α=sin1∘1​∑k=029​[cot(60+2k)∘−cot(61+2k)∘]

  4. Evaluate the sum: Let S be the sum part: S=∑k=029[cot⁡(60+2k)∘−cot⁡(61+2k)∘]S = \sum_{k=0}^{29} \left[ \cot(60+2k)^{\circ} - \cot(61+2k)^{\circ} \right]S=∑k=029​[cot(60+2k)∘−cot(61+2k)∘] Expanding the sum, we get: S=(cot⁡60∘−cot⁡61∘)+(cot⁡62∘−cot⁡63∘)+⋯+(cot⁡118∘−cot⁡119∘)S = (\cot 60^{\circ} - \cot 61^{\circ}) + (\cot 62^{\circ} - \cot 63^{\circ}) + \cdots + (\cot 118^{\circ} - \cot 119^{\circ})S=(cot60∘−cot61∘)+(cot62∘−cot63∘)+⋯+(cot118∘−cot119∘) This is not a direct telescoping series. Let's rearrange the terms by grouping the positive and negative terms: S=(cot⁡60∘+cot⁡62∘+⋯+cot⁡118∘)−(cot⁡61∘+cot⁡63∘+⋯+cot⁡119∘)S = (\cot 60^{\circ} + \cot 62^{\circ} + \cdots + \cot 118^{\circ}) - (\cot 61^{\circ} + \cot 63^{\circ} + \cdots + \cot 119^{\circ})S=(cot60∘+cot62∘+⋯+cot118∘)−(cot61∘+cot63∘+⋯+cot119∘) Let Seven=cot⁡60∘+cot⁡62∘+⋯+cot⁡118∘S_{even} = \cot 60^{\circ} + \cot 62^{\circ} + \cdots + \cot 118^{\circ}Seven​=cot60∘+cot62∘+⋯+cot118∘. Let Sodd=cot⁡61∘+cot⁡63∘+⋯+cot⁡119∘S_{odd} = \cot 61^{\circ} + \cot 63^{\circ} + \cdots + \cot 119^{\circ}Sodd​=cot61∘+cot63∘+⋯+cot119∘.

  5. Simplify SevenS_{even}Seven​ and SoddS_{odd}Sodd​: We use the identity cot⁡(180∘−x)=−cot⁡x\cot(180^{\circ} - x) = -\cot xcot(180∘−x)=−cotx. For SevenS_{even}Seven​: Seven=(cot⁡60∘+cot⁡62∘+⋯+cot⁡88∘)+cot⁡90∘+(cot⁡92∘+⋯+cot⁡118∘)S_{even} = (\cot 60^{\circ} + \cot 62^{\circ} + \cdots + \cot 88^{\circ}) + \cot 90^{\circ} + (\cot 92^{\circ} + \cdots + \cot 118^{\circ})Seven​=(cot60∘+cot62∘+⋯+cot88∘)+cot90∘+(cot92∘+⋯+cot118∘) Using the identity, cot⁡118∘=−cot⁡62∘\cot 118^{\circ} = -\cot 62^{\circ}cot118∘=−cot62∘, cot⁡116∘=−cot⁡64∘\cot 116^{\circ} = -\cot 64^{\circ}cot116∘=−cot64∘, ..., cot⁡92∘=−cot⁡88∘\cot 92^{\circ} = -\cot 88^{\circ}cot92∘=−cot88∘. Also, cot⁡90∘=0\cot 90^{\circ} = 0cot90∘=0. Seven=cot⁡60∘+(cot⁡62∘−cot⁡62∘)+(cot⁡64∘−cot⁡64∘)+⋯+(cot⁡88∘−cot⁡88∘)+0S_{even} = \cot 60^{\circ} + (\cot 62^{\circ} - \cot 62^{\circ}) + (\cot 64^{\circ} - \cot 64^{\circ}) + \cdots + (\cot 88^{\circ} - \cot 88^{\circ}) + 0Seven​=cot60∘+(cot62∘−cot62∘)+(cot64∘−cot64∘)+⋯+(cot88∘−cot88∘)+0 Seven=cot⁡60∘S_{even} = \cot 60^{\circ}Seven​=cot60∘

    For SoddS_{odd}Sodd​: Sodd=(cot⁡61∘+cot⁡63∘+⋯+cot⁡89∘)+(cot⁡91∘+⋯+cot⁡119∘)S_{odd} = (\cot 61^{\circ} + \cot 63^{\circ} + \cdots + \cot 89^{\circ}) + (\cot 91^{\circ} + \cdots + \cot 119^{\circ})Sodd​=(cot61∘+cot63∘+⋯+cot89∘)+(cot91∘+⋯+cot119∘) Using the identity, cot⁡119∘=−cot⁡61∘\cot 119^{\circ} = -\cot 61^{\circ}cot119∘=−cot61∘, cot⁡117∘=−cot⁡63∘\cot 117^{\circ} = -\cot 63^{\circ}cot117∘=−cot63∘, ..., cot⁡91∘=−cot⁡89∘\cot 91^{\circ} = -\cot 89^{\circ}cot91∘=−cot89∘. Sodd=(cot⁡61∘−cot⁡61∘)+(cot⁡63∘−cot⁡63∘)+⋯+(cot⁡89∘−cot⁡89∘)S_{odd} = (\cot 61^{\circ} - \cot 61^{\circ}) + (\cot 63^{\circ} - \cot 63^{\circ}) + \cdots + (\cot 89^{\circ} - \cot 89^{\circ})Sodd​=(cot61∘−cot61∘)+(cot63∘−cot63∘)+⋯+(cot89∘−cot89∘) Sodd=0S_{odd} = 0Sodd​=0

  6. Calculate α: Now we can find S: S=Seven−Sodd=cot⁡60∘−0=cot⁡60∘S = S_{even} - S_{odd} = \cot 60^{\circ} - 0 = \cot 60^{\circ}S=Seven​−Sodd​=cot60∘−0=cot60∘ Substitute this back into the expression for α: α=1sin⁡1∘⋅S=cot⁡60∘sin⁡1∘\alpha = \frac{1}{\sin 1^{\circ}} \cdot S = \frac{\cot 60^{\circ}}{\sin 1^{\circ}}α=sin1∘1​⋅S=sin1∘cot60∘​

  7. Calculate the final value: We need to find the value of (cosec⁡1∘α)2\left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}\right)^2(αcosec1∘​)2. First, calculate the base of the expression: cosec⁡1∘α=1/sin⁡1∘cot⁡60∘/sin⁡1∘=1cot⁡60∘=tan⁡60∘\frac{\operatorname{cosec} 1^{\circ}}{\alpha} = \frac{1/\sin 1^{\circ}}{\cot 60^{\circ}/\sin 1^{\circ}} = \frac{1}{\cot 60^{\circ}} = \tan 60^{\circ}αcosec1∘​=cot60∘/sin1∘1/sin1∘​=cot60∘1​=tan60∘ Now, square this value: (cosec⁡1∘α)2=(tan⁡60∘)2=(3)2=3\left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}\right)^2 = (\tan 60^{\circ})^2 = (\sqrt{3})^2 = 3(αcosec1∘​)2=(tan60∘)2=(3​)2=3

Final Answer: The value of the expression is 3.

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