Question

Solve this:-

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

Answer 1

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 ∅

____________________

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

____________________

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

____________________

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

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

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

____________________

I hope it helps you ❤️✔️

Answer 2

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

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