Question

prove that 1-cosA / 1+cosA=( cotA- cosecA)^2​

Answer 1

To Prove :-

$\sf \dfrac{1-cosA}{1+cosA} = (cotA- cosecA)^2$

Solution :-

$\sf L.H.S = \dfrac{1-cosA}{1+cosA}$

Multiply numerator and denominator by $\sf 1- cosA$,

       $= \sf \dfrac{(1-cosA)\times (1-cosA} {(1+cosA)\times (1-cosA)} \\\\= \dfrac{(1-cosA)^2}{1-cos^2A}$

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

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

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

        $= \sf (cosecA-cotA)^2 \\\\= (cotA-cosecA)^2 \\\\= R.H.S$

Hence proved

Answer 2

Question$\frac{1-cosA}{1+cosaA}$ = (cotA- cosecA)²solution LHS$\frac{1-cosA}{1+cosaA}$ (Multiply numerator and denominator by 1- cosA)= [tex]\frac{(

[tex]\frac{1-cos