Prove the following trigonometric identity: $cot (45° -A)$ = $tan 2A + sec 2A$
LHS
= $cot (45° -A)$
= $\dfrac{cot45°.cotA + 1}{cotA - cot45°}$
= $\dfrac{cot A + 1}{cot A - 1}$
= $\dfrac{\frac{cosA}{sinA} +1}{\frac{cosA}{sinA} -1}$
= $\dfrac{\frac{cosA +sinA}{sinA}}{\frac{cosA -sinA}{sinA}}$
= $\dfrac{cosA +sinA}{cosA -sinA}$
= $\dfrac{(cosA +sinA)(cosA +sinA)}{(cosA -sinA)(cosA +sinA)}$
= $\dfrac{cos²A +sin²A + 2sinAcosA}{cos²A - sin²A}$
= $\dfrac{1 +sin2A}{cos2A}$
= $\dfrac{1}{cos2A} + \dfrac{sin2A}{cos2A}$
= $sec2A + tan2A$
= $tan 2A + sec 2A$
= RHS
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