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If θ is an Acute Angle Such that Tan 2 θ = 8 7 Then the Value of ( 1 + Sin θ ) ( 1 − Sin θ ) ( 1 + Cos θ ) ( 1 − Cos θ ) - Mathematics

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Question

If θ is an acute angle such that \[\tan^2 \theta = \frac{8}{7}\] then the value of \[\frac{\left( 1 + \sin \theta \right) \left( 1 - \sin \theta \right)}{\left( 1 + \cos \theta \right) \left( 1 - \cos \theta \right)}\]

Options

  • \[\frac{7}{8}\]

  • \[\frac{8}{7}\]

  • \[\frac{7}{4}\]

  • \[\frac{64}{49}\]

MCQ

Solution

Given that:  `tan^2 θ=8/7` and θis an acute angle

We have to find the following expression  `((1+sinθ)(1-sin θ))/((1+cos θ)(1-cos θ))`

Since 

`tan^2θ=8/7` 

`tan θ=sqrt(8/7)` 

`tan θ=sqrt8/sqrt7` 

Since `tan θ="perpendiular"/"Base"` 

⇒ `"Perpendicular"=sqrt8` 

⇒ `"Base"=sqrt7` 

⇒ `"Hypotenuse"
= sqrt(8+7)` 

⇒ `"Hypotenuse"=sqrt15` 

We know that  `sinθ= "Perpendicular"/"Hypotenuse" and cos θ="Base"/"Hypotenuse"` 

We find:

`((1+sinθ )(1-sin θ))/((1+cos θ)(1-cosθ))` 

=`((1+sqrt8/sqrt15)(1-sqrt8/sqrt15))/((1+sqrt7/sqrt15)(1-sqrt7/sqrt15))`

=`((1-8/15))/((1-7/15))`

=`(7/15)/(8/15)` 

=`7/8` 

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Chapter 10: Trigonometric Ratios - Exercise 10.5 [Page 57]

APPEARS IN

RD Sharma Mathematics [English] Class 10
Chapter 10 Trigonometric Ratios
Exercise 10.5 | Q 8 | Page 57

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