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Trigonometric Ratio and Identites question

2024 · 8 Apr · Shift 1 · Q46
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Trigonometric Ratio and Identites question

2024 · 8 Apr · Shift 1 · Q46

JEE MainMathematicsTrigonometric Ratio and IdentitesMCQ+4 / −1
If sin⁡x=−35\sin x=-\frac{3}{5}sinx=−53​, where π<x<3π2\pi\lt x \lt \frac{3 \pi}{2}π<x<23π​, then 80(tan⁡2x−cos⁡x)80\left(\tan ^2 x-\cos x\right)80(tan2x−cosx) is equal to
  1. A
    109
  2. B
    108
  3. C
    19
  4. D
    18
View written solutionFree

Correct answer: A

  1. Identify the quadrant

Given: sin⁡x=−35,π<x<3π2\sin x=-\frac{3}{5}, \qquad \pi < x < \frac{3\pi}{2}sinx=−53​,π<x<23π​

The interval (π,3π2)\left(\pi,\frac{3\pi}{2}\right)(π,23π​) is the third quadrant.

In the third quadrant:

  • sin⁡x<0\sin x<0sinx<0
  • cos⁡x<0\cos x<0cosx<0
  • tan⁡x>0\tan x>0tanx>0
  1. Find cos⁡x\cos xcosx

Using: sin⁡2x+cos⁡2x=1\sin^2 x+\cos^2 x=1sin2x+cos2x=1

We have: cos⁡2x=1−sin⁡2x=1−(35)2=1−925=1625\cos^2 x=1-\sin^2 x=1-\left(\frac{3}{5}\right)^2=1-\frac{9}{25}=\frac{16}{25}cos2x=1−sin2x=1−(53​)2=1−259​=2516​

So: cos⁡x=±45\cos x=\pm \frac{4}{5}cosx=±54​

Since xxx is in the third quadrant, cos⁡x<0\cos x<0cosx<0, hence: cos⁡x=−45\cos x=-\frac{4}{5}cosx=−54​

  1. Find tan⁡2x\tan^2 xtan2x

tan⁡x=sin⁡xcos⁡x=−3/5−4/5=34\tan x=\frac{\sin x}{\cos x}=\frac{-3/5}{-4/5}=\frac{3}{4}tanx=cosxsinx​=−4/5−3/5​=43​

Thus, tan⁡2x=(34)2=916\tan^2 x=\left(\frac{3}{4}\right)^2=\frac{9}{16}tan2x=(43​)2=169​

  1. Compute the expression

We need: 80(tan⁡2x−cos⁡x)80\left(\tan^2 x-\cos x\right)80(tan2x−cosx)

Substitute the values: 80(916−(−45))=80(916+45)80\left(\frac{9}{16}-\left(-\frac{4}{5}\right)\right)=80\left(\frac{9}{16}+\frac{4}{5}\right)80(169​−(−54​))=80(169​+54​)

Take LCM of 161616 and 555: 916+45=4580+6480=10980\frac{9}{16}+\frac{4}{5}=\frac{45}{80}+\frac{64}{80}=\frac{109}{80}169​+54​=8045​+8064​=80109​

Therefore, 80⋅10980=10980\cdot \frac{109}{80}=10980⋅80109​=109

  1. Compare with options

The value is: 109\boxed{109}109​

So the correct option is A.

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