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If 160 3 Tan θ = a B , Then a Sin θ + B Cos θ a Sin θ − B Cos θ - Mathematics

MCQ

If \[\frac{160}{3}\] \[\tan \theta = \frac{a}{b}, \text{ then } \frac{a \sin \theta + b \cos \theta}{a \sin \theta - b \cos \theta}\]

 

Options

  • \[\frac{a^2 + b^2}{a^2 - b^2}\]

  • \[\frac{a^2 - b^2}{a^2 + b^2}\]

  • \[\frac{a + b}{a - b}\]

  • \[\frac{a - b}{a + b}\]

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Solution

Given :` tan θ = a/b' 

We have to find the value of following expression in terms of a and b

We know that:  `tanθ="Perpendicular"/"Base"`

⇒`" Base" =b` 

⇒` "Perpendicular=a"` 

⇒`" Hypotenuse"=sqrt (a^2+b^2)` 

Now we find,

`(a sinθ+b cos θ)/(a sinθ-b cos θ)=(a(a/(a^2+b^2))+b (b/(a^2+b^2)))/(a(a/(a^2+b^2))-b(b/(a^2+b^2)))` 

=`((a^2+b^2)/(a^2+b^2))/((a^2-b^2)/(a^2+b^2))`

=`(a^2+b^2)/(a^2-b^2)`

Hence the correct option is  (a)

 

 

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APPEARS IN

RD Sharma Class 10 Maths
Chapter 10 Trigonometric Ratios
Q 2 | Page 56
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