Math, asked by 2007Adibhai9th, 11 hours ago

Answer my last three questions which are of mathematics.​

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Answered by IamOnePunchMan
0

Answer:

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

Consider LHS

\rm \: sinA(secA + cosecA) - cosA(secA - cosecA)

We know,

\boxed{\tt{  \: secx =  \frac{1}{cosx} \: }} \\

and

\boxed{\tt{  \: cosecx =  \frac{1}{sinx} \: }} \\

So, using this identity, we can rewrite as

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

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

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

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

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

\rm \:  =  \: \dfrac{ 1}{sinA \: cosA}

\rm \:  =  \: secA \: cosecA

Hence Proved

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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

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