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If Tan θ = 1 √ 7 , Then C O S E C 2 θ − Sec 2 θ C O S E C 2 θ + Sec 2 θ =V - Mathematics

MCQ

If \[\tan \theta = \frac{1}{\sqrt{7}}, \text{ then } \frac{{cosec}^2 \theta - \sec^2 \theta}{{cosec}^2 \theta + \sec^2 \theta} =\] 

Options

  • \[\frac{5}{7}\]

  • \[\frac{3}{7}\]

  • \[\frac{1}{12}\]

  • \[\frac{3}{4}\]

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Solution

Given that:

`tan θ=1/sqrt7`

We are asked to find the value of the following expression

`(cosec^2θ-sec^2θ)/(cosec^2θ+sec^2θ)`

Since `tan θ= "Perpendicular"/"Base"` .

⇒ `"Perpendicular"=1` 

⇒ `"Base"= sqrt7` 

⇒ `"Hypotenuse"=sqrt(1+7)` 

⇒`" Hypotenuse"=sqrt8`

We know that `secθ="Hypotenuse"/"Base"  and  cosecθ= "Hypotenuse"/"Perpendicular"` 

We find:

`(Cosec^2θ-sec^2 θ)/(Cosec^2 +sec^2 θ)`

`((sqrt8/1)^2-(sqrt8/sqrt7)^2)/((sqrt8/1)^2+(sqrt8/sqrt7)^2)`

=(8/1-8/7)/(8/1+8/7)

=`(48/7)/(64/7)`

=`3/4`

 

 

 

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APPEARS IN

RD Sharma Class 10 Maths
Chapter 10 Trigonometric Ratios
Q 6 | Page 56
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