Math, asked by nancy359, 3 months ago

Solve this:-

 \frac{1 + tan {}^{2}A }{1 +  {cot}^{2} A}

Answers

Answered by kinzal
4

Explanation :-

 \sf \longrightarrow \frac{1 + tan^{2}A }{1 + {cot}^{2} A} \\

 \sf \longrightarrow \frac{sec² A }{cosec² A} \\

 \sf \longrightarrow \frac{\frac{1}{cos²A}}{\frac{1}{sin²A}} \\

 \sf \longrightarrow \frac{1}{cos²A} × \frac{sin²A }{1} \\

 \sf \longrightarrow \frac{sin²A}{cos² A} \\

 \sf \longrightarrow \underline{ \underline{tan² A} } \\

For More Information :-

  • Sin² ∅ + Cos² = 1
  • Sin² ∅ = 1 - Cos² ∅
  • Cos² ∅ = 1 - Sin² ∅

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  • Sec² ∅ - Tan² ∅ = 1
  • Sec² ∅ = 1 + Tan² ∅
  • Tan² ∅ = 1 + Sec² ∅

____________________

  • Cosec² ∅ - Cot² ∅ = 1
  • Cosec² ∅ = 1 + Cot² ∅
  • Cot² ∅ = 1 + Cosec² ∅

____________________

  •  \sf \longrightarrow \frac{1}{Sin \: \: A} = Cosec \: \:  A \\

  •  \sf \longrightarrow \frac{1}{Cos \: \: A} = Sec \: \:  A \\

  •  \sf \longrightarrow \frac{1}{Tan \: \: A} = Cot \: \:  A \\

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I hope it helps you ❤️✔️

Answered by mathdude500
5

\large\underline{\sf{Solution-}}

Given Trigonometric function is

 \red{\rm :\longmapsto\:\dfrac{1 +  {tan}^{2} A}{1 +  {cot}^{2} A} }

We know,

 \purple{\boxed{ \tt{ \: cotx =  \frac{1}{tanx} \: }}}

So, using this identity, we get

 \red{\rm \:  =  \:\dfrac{1 +  {tan}^{2}A }{ \:  \:  \: 1 + \dfrac{1}{ {tan}^{2} A} }  \:  \:  \:  \:  \:  \:  \: }

 \red{\rm \:  =  \:\dfrac{1 +  {tan}^{2}A }{ \:  \:  \: \dfrac{{tan}^{2} A + 1}{ {tan}^{2} A} }  \:  \:  \:  \:  \:  \:  \: }

 \red{\rm \:  =  \: \cancel{(1 + {tan}^{2} A) }\times \dfrac{{tan}^{2} A}{\cancel{1 + {tan}^{2} A}} }

 \red{\rm \:  =  \:{tan}^{2} A}

Hence,

 \red{\rm \implies\:\boxed{ \tt{ \: \:\dfrac{1 +  {tan}^{2} A}{1 +  {cot}^{2} A} } = {tan}^{2} A}}

Alternative Method :-

Given Trigonometric function is

 \green{\rm :\longmapsto\:\dfrac{1 +  {tan}^{2} A}{1 +  {cot}^{2} A} }

We know,

 \purple{\boxed{ \tt{ \: tanx =  \frac{sinx}{cosx} \:  \: and \:  \: cotx =  \frac{cosx}{sinx} \:  \: }}}

So, using these Identities, we get

 \green{ \rm =  \:\dfrac{1 + \dfrac{ {sin}^{2} A}{ {cos}^{2} A} }{1 + \dfrac{ {cos}^{2} A}{ {sin}^{2} A} }  }

 \green{ \rm =  \:\dfrac{\dfrac{ {cos}^{2}A +   {sin}^{2} A}{ {cos}^{2} A} }{ \dfrac{ {sin}^{2}A +  {cos}^{2} A}{ {sin}^{2} A} }  }

\green{ \rm =  \:\dfrac{ {sin}^{2} A}{ {cos}^{2} A}}

\green{ \rm =   \:  {tan}^{2} A}

Hence,

 \green{\rm \implies\:\boxed{ \tt{ \: \:\dfrac{1 +  {tan}^{2} A}{1 +  {cot}^{2} A} } = {tan}^{2} A}}

Alternative Method :-

Given Trigonometric function is

 \pink{\rm :\longmapsto\:\dfrac{1 +  {tan}^{2} A}{1 +  {cot}^{2} A} }

\pink{ \rm =  \:\dfrac{ {sec}^{2} A}{ {cosec}^{2} A} }

\pink{ \rm =  \:\dfrac{\dfrac{1}{ {cos}^{2} A} }{\dfrac{1}{ {sin}^{2} A} } }

\pink{ \rm =  \:\dfrac{ {sin}^{2} A}{ {cos}^{2} A}}

\pink{ \rm =  \: {tan}^{2} A}

Hence,

 \pink{\rm \implies\:\boxed{ \tt{ \: \:\dfrac{1 +  {tan}^{2} A}{1 +  {cot}^{2} A} } = {tan}^{2} A}}

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

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