Question

prove that cos θ /1-tanθ +sinθ/1-cotθ =cosθ+sinθ​

Answer 1

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \red{\bf{tanx = \dfrac{sinx}{cosx}}}}$

$\boxed{ \red{\bf{cotx = \dfrac{cosx}{sinx}}}}$

$\boxed{ \red{\bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}}}$

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

Consider,

$\rm :\longmapsto\:\dfrac{cos\theta}{1 - tan\theta} + \dfrac{sin\theta}{1 - cot\theta}$

$\rm \: \: = \: \: \dfrac{cos\theta}{ \: \: \: 1 - \dfrac{sin\theta}{cos\theta} \: \: \: \: } + \dfrac{sin\theta}{ \: \: \: 1 - \dfrac{cos\theta}{sin\theta} \: \: \: }$

$\rm \: \: = \: \: \dfrac{cos\theta}{ \: \: \: \dfrac{cos\theta - sin\theta}{cos\theta} \: \: \: \: } + \dfrac{sin\theta}{ \: \: \: \dfrac{sin\theta - cos\theta}{sin\theta} \: \: \: }$

$\rm \: \: = \: \: \dfrac{ {cos}^{2}\theta }{cos\theta - sin\theta} + \dfrac{ {sin}^{2}\theta }{sin\theta - cos\theta}$

$\rm \: \: = \: \: \dfrac{ {cos}^{2}\theta }{cos\theta - sin\theta} - \dfrac{ {sin}^{2}\theta }{cos\theta - sin\theta}$

$\rm \: \: = \: \: \dfrac{ {cos}^{2}\theta - {sin}^{2}\theta }{cos\theta - sin\theta}$

$\rm \: \: = \: \: \dfrac{(cos\theta + sin\theta) \: \cancel{(cos\theta - sin\theta)}}{ \cancel{(cos\theta - sin\theta)}}$

$\rm \: \: = \: \: cos\theta + sin\theta$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

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