ত্রিকোমিতি ৯.১ এর ৬(গ) সমাধান।
- গ) $\displaystyle \frac{1}{sin^{2} A} -\frac{1}{tan^{2} A} =1$
L.H.S= $\displaystyle \frac{1}{sin^{2} A} -\frac{1}{tan^{2} A}$
$\displaystyle =cosec^{2} A-cot^{2} A$ [ We know that, $\displaystyle cosec^{2} A=\frac{1}{sin^{2} A} \ or,cot^{2} A=\frac{1}{tan^{2} A} \ $ ]
$\displaystyle =1$ [ Again, we knoe that: $\displaystyle cosec^{2} A-cot^{2} A$
=R.H.S
so, $\displaystyle \frac{1}{sin^{2} A} -\frac{1}{tan^{2} A} =1$ ( Proved )
Alternative solution:
গ) $\displaystyle \frac{1}{sin^{2} A} -\frac{1}{tan^{2} A} =1$
L.H.S= $\displaystyle \frac{1}{sin^{2} A} -\frac{1}{tan^{2} A}$
$\displaystyle =\frac{1}{sin^{2} A} -\frac{1}{\frac{sin^{2} A}{cos^{2} A}}$ [ we know, $\displaystyle tan^{2} A=\frac{sin^{2} A}{cos^{2} A}$ ]
$\displaystyle =\frac{1}{sin^{2} A} -\frac{cos^{2} A}{sin^{2} A}$
$\displaystyle =\frac{1-cos^{2} A}{sin^{2} A}$ [ we know, $\displaystyle sin^{2} A+cos^{2} A=1\ or,\ sin^{2} A=1-cos^{2} A$ ]
$\displaystyle =\frac{sin^{2} A}{sin^{2} A}$
$\displaystyle =1$
=R.H.S
so, $\displaystyle \frac{1}{sin^{2} A} -\frac{1}{tan^{2} A} =1$ ( Proved )
