Question

$\bf{ \pink{Prove \: \: that:}}$

$\displaystyle{\sf\:\dfrac{\cos\:A}{1\:-\:\tan\:A}\:+\:\dfrac{\sin^2\:A}{\sin\:A\:-\:\cos\:A}\:=\:\sin\:A\:+\:\cos\:A} \ \textless \ br /\ \textgreater$

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(with Step by step Explanation)​

Answer 1

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

Consider LHS

$\rm \: \dfrac{cosA}{1 - tanA} + \dfrac{ {sin}^{2} A}{sinA - cosA}$

We know,

$\boxed{\sf{ \:tanA = \frac{sinA}{cosA} \: }}$

So, on substituting, we get

$\rm \: = \: \dfrac{cosA}{1 - \dfrac{sinA}{cosA} } + \dfrac{ {sin}^{2}A}{sinA - cosA}$

$\rm \: = \: \dfrac{cosA}{\dfrac{cosA - sinA}{cosA} } + \dfrac{ {sin}^{2}A}{sinA - cosA}$

$\rm \: = \: \dfrac{cos^{2} A}{cosA - sinA} + \dfrac{ {sin}^{2}A}{sinA - cosA}$

$\rm \: = \: \dfrac{cos^{2} A}{cosA - sinA} - \dfrac{ {sin}^{2}A}{cosA - sinA}$

$\rm \: = \: \dfrac{cos^{2} A - {sin}^{2}A}{cosA - sinA}$

We know,

$\boxed{\sf{ \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: }}$

So, using this identity, we get

$\rm \: = \: \dfrac{(cosA - sinA)(cosA + sinA)}{cosA - sinA}$

$\rm \: = \: cosA + sinA$

Hence,

$\rm \: \dfrac{cosA}{1 - tanA} + \dfrac{ {sin}^{2} A}{sinA - cosA} = cosA + sinA$

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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

Answer 2

Step-by-step explanation:

We have,

$\pink{\begin{aligned} \text { LHS } & \tt=\frac{\cos A }{1-\tan A }+\frac{\sin ^{2} A }{\sin A -\cos A } \\ \\ & \tt=\frac{\cos A }{1-\frac{\sin A }{\cos A }+\frac{\sin ^{2} A }{\sin A -\cos A }} \\ \\ & \tt=\frac{\cos A }{\frac{\cos A -\sin A }{\cos A }+\frac{\sin ^{2} A }{\sin A -\cos A }} \\ \\ & \tt=\frac{\cos ^{2} A }{\cos A -\sin A }+\frac{\sin ^{2} A }{\sin A -\cos A } \\ \\ & \tt=\frac{\cos ^{2} A }{\cos A -\sin A }-\frac{\sin ^{2} A }{\cos A -\sin A } \\ \\ & \tt=\frac{\cos ^{2} A -\sin ^{2} A }{\cos A -\sin A } \\ \\ & \tt=\frac{(\cos A +\sin A )(\cos A -\sin A )}{\cos A -\sin A } \\ & =\cos A +\sin A = \text{RHS} \end{aligned} }$