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..​ Solve on paper​

Answer 1

Appropriate Question :-

Prove that,

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

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

Consider LHS

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

We know,

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

and

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

So, using these Identities, we get

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

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

We know,

$\boxed{ \tt{ \: (x - y)( {x}^{2} + {y}^{2} + xy) = {x}^{3} - {y}^{3}\: \: }}$

So, using this, we get

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

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

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

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

Hence,

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

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

Answer 2

Answer:

Statement is wrong

Step-by-step explanation:

If we substitute x = 30, LHS is not equals to RHS