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If 16 Cot X = 12, Then Sin X − Cos X Sin X + Cos X - Mathematics

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प्रश्न

If 16 cot x = 12, then \[\frac{\sin x - \cos x}{\sin x + \cos x}\]

पर्याय

  • \[\frac{1}{7}\]

  • \[\frac{3}{7}\]

  • \[\frac{2}{7}\]

  • 0

MCQ

उत्तर

We are given`16 cot x=12` .We are asked to find the following

`(sin x-cos x)/(sin x+cos x)`

We know that: `cot x= "Base"/"Perpendicular" `

⇒ "Base"=3

⇒ "Perpendicular"=4

⇒ `"Hypotenuse"= sqrt(("Perpendicular")^2+("Base")^2)`

⇒ `"Hypotenuse"=sqrt(16+9)`

⇒`"Hypotenuse"=5`

Now we have

`16 cot x=12`

`cot x=12/16`

`cot x=3/4`,

We know sin x=`"Perpendicular"/"Hypotenuse" and Cos x= "Base"/"Hypotenuse"`

Now we find

`(Sin x- cos x)/(sin z+cos x)`

= `(4/5-3/5)/(4/5+3/5)`

=`(1/5)/(7/5)`

=`1/7`

 

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पाठ 10: Trigonometric Ratios - Exercise 10.5 [पृष्ठ ५६]

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आरडी शर्मा Mathematics [English] Class 10
पाठ 10 Trigonometric Ratios
Exercise 10.5 | Q 4 | पृष्ठ ५६

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