Question

Prove that SinA (sec A + cosec A) -cos A (sec A- cosec A) = sec A cosec A.​

Answer 1

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

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

Answer 2

$\sin(a) ( \sec \: a + \cosec \: a) - \cos \: a( \sec \: a - cosec \: a) \\ we \: know \: that \: \\ sec \: a = \frac{1}{cos \: a} \\ and \\ cosec \: a = \frac{1}{sina} \\ sin \: a( \frac{1}{cos \: a} + \frac{1}{sin \: a} ) - cos \: a( \frac{1}{cos \: \: a} - \frac{1}{sin\: a} ) \\ sin \: a( \frac{sin \: a + cos \: a}{sina \: cosa} ) - cos \: a( \frac{sin a\: - \: cos \: a}{sina \: cosa} ) \\ \frac{sin {}^{2}a + sinacosa}{sinacosa} - \frac{sinacosa + cos {}^{2} \: a }{sinacosa} \\ \frac{sin {}^{2}a + sinacosa - sinacosa + cos {}^{2} a }{sinacosa} \\ \frac{sin {}^{2} a + cos {}^{2} a}{sinacosa} \\ by \: using \: identy \\ sin {}^{2} + \cos {}^{2} a = 1 \\ \frac{1}{sinacosa} \\ cosec \: a \: sec \: a$