Question

Prove that (1+cotA-cosecA) (1+TanA+secA)=2​

Answer 1

TO PROVE THAT :-

$\sf (1+cotA-cosecA)(1+tanA+secA) = 2$

SOLUTION :-

$\sf L.H.S = (1+cotA-cosecA)(1+tanA+secA)$

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

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

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

         $=\sf \dfrac{1-1+2sinAcosA}{sinAcosA} \\\\= \dfrac{2sinAcosA}{sinAcosA} \\\\= 2 \\\\= R.H.S$

Hence proved.