Question

Prove that (CosecA-Sin A) (SecA - Cos A)=1/tan A + cota​

Answer 1

Prove That:

$(CosecA - SinA)(SecA - cosA) = \frac{1}{tanA \: + \: cotA}$

Proof:

LHS Simplification:

$= (CosecA - SinA)(SecA - cosA) \\\\= (\frac{1}{sinA} - sinA)(\frac{1}{cosA} - cosA)\\\\= (\frac{1 - sin^{2}A}{sinA})(\frac{1 - cos^{2}A }{cosA} )\\\\= \frac{cos^{2}A }{sinA} \times \frac{sin^{2}A}{cosA} \\\\= cosA \times sinA \\\\= sinA.CosA$

RHS Simplification:

$=\frac{1}{tanA \: + \: cotA} \\\\=\frac{1}{\frac{sinA}{cosA} \: + \: \frac{cosA}{sinA} } \\\\=\frac{1}{\frac{sin^{2}A \: + \: cos^{2}A}{sinA.cosA} }\\\\=\frac{1}{\frac{1}{sinA.cosA} } \\\\= sinA.cosA$

LHS = RHS.

SinA.CosA = SinA.CosA.

Hence Proved...

Trigonometric Identities Used:

$1. \: CosecA = \frac{1}{SinA} \\\\2. \: SecA = \frac{1}{CosA} \\\\3. \: Sin^{2}A + Cos^{2}A = 1$

$\\\\ 4. \: tanA = \frac{sinA}{cosA}\\\\5. \: cotA = \frac{cosA}{sinA}$

Answer 2

Answer:

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