Question

Prove that sinA/ (1+cot A) - COSA/ (1+tana)
=
sinA-COSA.​

Answer 1

$\large\underline{ \sf{Given \:Question - }}$

Prove that,

$\rm :\longmapsto\:\dfrac{sinA}{1 + cotA} - \dfrac{cosA}{1 + tanA} = sinA \: - \: cosA$

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

Consider, LHS

$\rm :\longmapsto\:\dfrac{sinA}{1 + cotA} - \dfrac{cosA}{1 + tanA}$

We know,

$\boxed{ \tt{ \: cotx = \frac{cosx}{sinx} \: }}$

and

$\boxed{ \tt{ \: tanx = \frac{sinx}{cosx} \: }}$

So, on substituting these values, we get

$\rm \: = \:\dfrac{sinA}{1 + \dfrac{cosA}{sinA} } - \dfrac{cosA}{1 + \dfrac{sinA}{cosA} }$

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

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

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

We know,

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

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

$\rm \: = \:sinA - cosA$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{sinA}{1 + cotA} - \dfrac{cosA}{1 + tanA} = sinA-cosA}}$

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

Answer:

$\sf\tt\large{\green {\underline {\underline{⚘\;Question :}}}}$

$\frac{sin \: A}{1 + cot \: A} - \frac{cosA }{1 + tan \: A } = sin \: A - cos \: A$

$\sf\tt\large{\green {\underline {\underline{⚘\;Solution :}}}}$

• Let,

• Consider, LHS now,

• $\frac{sin \: A}{1 + cot \: A} - \frac{cos \: A}{1 + tan \: A}$

Now,

• Substituting the values we get that,

• $= \frac{sin \: A}{1 + \frac{cos \: A}{sin \: A} } - \frac{cos \: A}{1 + \frac{sin \: A}{cos \: A} }$

• $= \frac{sin \: A}{ \frac{sin \: A + cos \: A}{sin \: A} } - \frac{cos \: A}{ \frac{cos \: A + sin \: A}{cos \: A} }$

• By solving this we get that,

• $= \frac{ {cos}^{2} A}{cos \: A + sin \: A} - \frac{ {sin}^{2} A}{cos \: A + sin \: A}$

• $= \frac{ {cos \: }^{2} A - {sin}^{2} A}{cos \: A + sin \: A}$

Now ,

• Applying the formula that is

${x}^{2} - {y}^{2} = (x + y)(x - y)$

• From this we get that,

• Sin A-cosA

Therefore ,

• LHS=RHS

Hope it helps u mate .

Thank you .