Question

26. Prove that
Sinθ
/1 - cotθ4 +
cos
/1 - tanθ
Sinθ+cosθ .​

Answer 1

Appropriate Question

Prove that

$\sf\:\dfrac{sin\theta}{1 - cot\theta} + \dfrac{cos\theta}{1 - tan\theta} = sin\theta + cos\theta$

Identities Used :-

$\red{\boxed{ \sf \: tan\theta = \dfrac{sin\theta}{cos\theta}}}$

$\red{\boxed{ \sf \: cot\theta = \dfrac{cos\theta}{sin\theta}}}$

Solution :-

Consider,

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

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

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

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

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

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

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

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

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

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