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NDA 2023
Exam
NDA
Year
2023
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2023 · Mathematics · Trigonometric Ratios & Identities

Paragraph: Given that x+ x+ x+ x+ x+cosec x=7 Question: The given...

Paragraph: Given that sin⁡x+cos⁡x+tan⁡x+cot⁡x+sec⁡x+cosec⁡x=7\sin x+\cos x+\tan x+\cot x+\sec x+\operatorname{cosec} x=7
Question: The given equation can be reduced to

Options

  1. Asin⁡22x−44sin⁡2x+36=0\sin ^2 2 x-44 \sin 2 x+36=0
  2. Bsin⁡22x+44sin⁡2x−36=0\sin ^2 2 x+44 \sin 2 x-36=0
  3. Csin⁡22x−22sin⁡2x+18=0\sin ^2 2 x-22 \sin 2 x+18=0
  4. Dsin⁡22x−22sin⁡2x+18=0\sin ^2 2 x-22 \sin 2 x+18=0

Correct answer

A. sin⁡22x−44sin⁡2x+36=0\sin ^2 2 x-44 \sin 2 x+36=0

Step-by-step solution

sin⁡x+cos⁡x+tan⁡x+cot⁡x+sec⁡x+cosec⁡x=7(sin⁡x+cos⁡x)+(sin⁡xcos⁡x+cos⁡xsin⁡x)+(1sin⁡x+1cos⁡x)=7(sin⁡x+cos⁡x)+1sin⁡xcos⁡x+sin⁡x+cos⁡xsin⁡xcos⁡x=7(sin⁡x+cos⁡x)×(1+1sin⁡xcos⁡x)+1sin⁡xcos⁡x=7(sin⁡x+cos⁡x)(sin⁡xcos⁡x+1)+1sin⁡xcos⁡x=7(sin⁡x+cos⁡x)(sin⁡xcos⁡x+1)+1=7sin⁡xcos⁡x(sin⁡x+cos⁡x)(sin⁡xcos⁡x+1)=7sin⁡xcos⁡x−1\sin x+\cos x+\tan x+\cot x+\sec x+\operatorname{cosec} x=7(\sin x+\cos x)+\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)+\left(\frac{1}{\sin x}+\frac{1}{\cos x}\right)=7(\sin x+\cos x)+\frac{1}{\sin x \cos x}+\frac{\sin x+\cos x}{\sin x \cos x}=7(\sin x+\cos x) \times\left(1+\frac{1}{\sin x \cos x}\right)+\frac{1}{\sin x \cos x}=7\frac{(\sin x+\cos x)(\sin x \cos x+1)+1}{\sin x \cos x}=7(\sin x+\cos x)(\sin x \cos x+1)+1=7 \sin x \cos x(\sin x+\cos x)(\sin x \cos x+1)=7 \sin x \cos x-1 Squaring both sides (1+sin⁡2x)(14sin⁡22x+1+sin⁡2x)=494sin⁡22x+1−7sin⁡2x\begin{aligned}(1+\sin 2 x)\left(\frac{1}{4} \sin ^2 2 x+1+\sin 2 x\right) & \\ & =\frac{49}{4} \sin ^2 2 x+1-7 \sin 2 x\end{aligned}(1+sin⁡2x)(sin⁡22x+4sin⁡2x+4)=49sin⁡22x−28sin⁡2x\begin{aligned}(1+\sin 2 x)\left(\sin ^2 2 x+4 \sin 2 x+4\right) & \\ & =49 \sin ^2 2 x-28 \sin 2 x\end{aligned}⇒sin⁡22x+4sin⁡2x+4+sin⁡32x+4sin⁡22x+4sin⁡2x=49sin⁡22x−28sin⁡2x+4\begin{array}{r}\Rightarrow \sin ^2 2 x+4 \sin 2 x+4+\sin ^3 2 x+4 \sin ^2 2 x+4 \sin 2 x \\ =49 \sin ^2 2 x-28 \sin 2 x+4\end{array}\Rightarrow \sin ^3 2 x-44 \sin ^2 2 x+36 \sin 2 x=0\Rightarrow \quad \sin 2 x\left(\sin ^2 2 x-44 \sin 2 x+36\right)=0\Rightarrow \sin ^2 2 x-44 \sin 2 x+36=0
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