Question

prove cotA-tanA=(2cos2A-1)/sinAcosA

Answer 1

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Answer 2

Solution:

LHS = cotA-tanA

= $\frac{cosA}{sinA}-\frac{sinA}{cosA}$

/* we know that,

i) $cotA = \frac{cosA}{sinA}$

ii) $tanA=\frac{sinA}{cosA}$ */

$\frac{cos^{2}A-sin^{2}A}{sinAcosA}$

= $\frac{cos^{2}A-(1-cos^{2}A}{sinAcosA}$

/* We know that,

i ) cos²A-sin²A = cos2A

ii ) sin²A = 1-cos²A */

= $\frac{cos^{2}A-1+cos^{2}A}{sinAcosA}$

= $\frac{2cos^{2}A-1}{sinAcosA}$

= RHS

Therefore,

$cotA-tanA=\frac{2cos^{2}A-1}{sinAcosA}$

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