ত্রিকোমিতি ৯.১ এর ৮(ক) সমাধান।
৮) (ক) $\displaystyle \frac{tanA}{1-cotA} +\frac{cotA}{1-tanA} =secA.cosecA+1$
L.H.S= $\displaystyle \frac{tanA}{1-cotA} +\frac{cotA}{1-tanA}$
$\displaystyle =\frac{\frac{sinA}{cosA}}{1-\frac{cosA}{sinA}} +\frac{\frac{cosA}{sinA}}{1-\frac{sinA}{cosA}}$
$\displaystyle =\frac{\frac{sinA}{cosA}}{\frac{sinA-cosA}{sinA}} +\frac{\frac{cosA}{sinA}}{\frac{cosA-sinA}{cosA}}$
$\displaystyle =\frac{sinA}{cosA} \ X\frac{sinA}{sinA-cosA} +\frac{cosA}{sinA} X\frac{cosA}{cosA-sinA}$
$\displaystyle =\frac{sin^{2} A}{cosA( sinA-cosA)} -\frac{cos^{2} A}{sinA( sinA-cosA)}$
$\displaystyle =\frac{sin^{3} A-cos^{3} A}{cosA.sinA( sinA-cosA)}$
$\displaystyle =\frac{( sinA-cosA)\left( sin^{2} A+sinA.cosA+cos^{2} A\right)}{cosA.sinA( sinA-cosA)}$
$\displaystyle =\frac{1+\ sinA.\ cosA}{sinA.cosA}$
$\displaystyle =\frac{1}{sinA.cosA} +\frac{sinA.cosA}{sinA.cosA}$
$\displaystyle =\frac{1}{sinA} .\frac{1}{cosA} +1$
$\displaystyle =cosecA.\ secA+1$
so, $\displaystyle \frac{tanA}{1-cotA} +\frac{cotA}{1-tanA} =secA.cosecA+1$ ( Proved).
