Question

prove that sec A minus Cos A cot A + tan a is equal to sec A tan A​

Answer 1

Step-by-step explanation:

LHS=(secA-cosA) * (cotA+tanA)

=(1/cosA - cosA) * cosA/sinA + sinA/cosA)

=(1-cos^2A)/cosA * (sin^2A + cos^2A ) / sinA cosA)

={(sin^2A) /cosA}*(1/sinA cosA)

=sinA/cos^2A = (sin A/cosA)*(1/cosA)

=tanA*secA=RHS

Answer 2

TO PROVE :-

$\sf (secA-cosA)\times (cotA+tanA)= secA \ tanA$

SOLUTION :-

$\sf L.H.S = (secA-cosA)\times (cotA+tanA)$

$\bullet\bf \ secA = \dfrac{1}{cosA} \\\\\bullet \ cotA = \dfrac{cosA}{sinA} \\\\\bullet \ tanA = \dfrac{sinA}{cosA}$

         $= \sf \left( \dfrac{1}{cosA} -cosA \right) \times \left( \dfrac{cosA}{sinA} + \dfrac{sinA}{cosA} \right) \\\\= \left( \dfrac{1-cos^2A}{cosA} \right) \times \left( \dfrac{cos^2A+sin^2A}{sinAcosA} \right)$

$\bullet\bf \ 1-cos^2A = sin^2A \\\\\bullet \ cos^2A+sin^2A = 1$

          $= \sf \dfrac{sin^2A}{cosA} \times \dfrac{1}{sinAcosA} \\\\= \dfrac{sinA}{cosA}\times \dfrac{1}{cosA} \\\\= tanA\times secA\\\\= secA \ tanA \\\\= R.H.S$

Hence proved.