Question

Only for Brainly moderators , Brainly stars and brainly best users ▬▬▬▬ஜ۩۞۩ஜ▬▬1 + cot A + tan a into Sin A minus Cos A is equal to sec a upon cos square A minus Cos A upon sec square A​

Answer 1

Answer:

L.H.S

=(1+ cos A /sinA + sinA/cosA)(sinA-CosA

=(cosAsinA + cos^2A + Sin^2A/cosAsinA )(sinA- cosA). (take LCM)

=(cosA sinA + 1/cosAsinA) (sin A- cos A ).(sin^2A +cos^2 A = 1)

= cosA sin^2A -cos^2 A sin A + sinA - cosA / cos A sin A  

=sinA ( 1 - cos^2 A) -cosA (1- sin^2A)/ cosAsinA

sinA sin^2A - cos A cos^2 /cosAsinA

sin^3A - cos^3A/cosAsinA

sin^3A/ sinAcosA - cos^3A/cosA sinA

sin^2A /cosA - cos^2 /sinA  

=secA/ cosecA - cosecA / sec^2.

HENCE PROVED  

L.H.S = R.H.S

Step-by-step explanation:

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Answer 2

Appropriate Question

$\rm\:(1 + tanA + cotA)(sinA - cosA) = \dfrac{secA}{ {cosec}^{2} A} - \dfrac{cosecA}{ {sec}^{2}A}$

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

The given Trigonometric function is

$\rm :\longmapsto\:(1 + tanA + cotA)(sinA - cosA) =$

We know,

$\red{\boxed{ \rm{ \: tanx = \frac{sinx}{cosx}}}} \: \: \: and \: \: \: \red{\boxed{ \rm{ \: cotx = \frac{cosx}{sinx}}}}$

On substituting these, we get

$\rm \: = \:(sinA - cosA)\bigg[1 + \dfrac{sinA}{cosA} + \dfrac{cosA}{sinA} \bigg]$

$\rm \: = \:(sinA - cosA)\bigg[\dfrac{sinAcosA + {sin}^{2}A + {cos}^{2}A}{sinAcosA} \bigg]$

can be re-arranged as

$\rm \: = \:(sinA - cosA)\bigg[\dfrac{{sin}^{2}A + {cos}^{2}A + sinAcosA}{sinAcosA} \bigg]$

We know,

$\red{\boxed{ \rm{ \: (x - y)( {x}^{2} + {y}^{2} + xy) \: }}}$

So, using this, we get

$\rm \: = \:\dfrac{ {sin}^{3}A - {cos}^{3}A}{sinAcosA}$

$\rm \: = \:\dfrac{ {sin}^{3}A}{sinAcosA} - \dfrac{ {cos}^{3}A}{sinAcosA}$

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

We know,

$\red{\boxed{ \rm{ \: sinx = \frac{1}{cosecx} \: }}} \: \: \: and \: \: \: \red{\boxed{ \rm{ \: cosx = \frac{1}{secx} \: }}}$

So, using these, we get

$\rm \: = \:\dfrac{secA}{ {cosec}^{2} A} - \dfrac{cosecA}{ {sec}^{2}A}$

Hence,

$\red{\boxed{ \rm{ \: \rm\:(1 + tanA + cotA)(sinA - cosA) = \dfrac{secA}{ {cosec}^{2} A} - \dfrac{cosecA}{ {sec}^{2}A}}}}$

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