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P Given 4 Cos θ − Sin θ 2 Cos θ + Sin θ What is the Value of C O S E C 2 θ − Sec 2 θ C O S E C 2 θ + Sec 2 θ - Mathematics

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प्रश्न

Given 

\[\frac{4 \cos \theta - \sin \theta}{2 \cos \theta + \sin \theta}\] what is the value of \[\frac{{cosec}^2 \theta - \sec^2 \theta}{{cosec}^2 \theta + \sec^2 \theta}\]

योग

उत्तर

Given: ` tan θ= 1/sqrt5`

We know that: `tan θ=("Prependicular")/("Base")` 

`("Prependicular")/("Base")=1/sqrt5`  

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

`"Hypotenuse"=sqrt(1+5)`

`"Hypotenuse"=sqrt6` 

Now we find, `(cosec^2θ-sec^2θ)/(cosec^2θ+sec^2θ)` 

=`(("hypotenuse")^2/("Perpendicular")^2-("hypotenuse")^2/("Base")^2)/(("hypotenuse")^2/("Perpendicular")^2+("hypotenuse")^2/("Base")^2)` 

= `((sqrt6)^2/(1)^2-(sqrt6)^2/(sqrt5)^2)/((sqrt6)^2/(1)^2+((sqrt6))/(sqrt5)^2)` 

= `(6/1-6/5)/(6/1+6/5)` 

=`(24/5)/(36/5)` 

=`2/3` 

Hence the value of `(cosec^2θ-sec^2θ)/(cosec^2θ+sec^2θ)` is `2/3` 

 

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अध्याय 10: Trigonometric Ratios - Exercise 10.4 [पृष्ठ ५५]

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आरडी शर्मा Mathematics [English] Class 10
अध्याय 10 Trigonometric Ratios
Exercise 10.4 | Q 8 | पृष्ठ ५५

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