amanf

2021-08-22

Prove that:
$\left(1+\frac{1}{{\mathrm{tan}}^{2}A}\right)\left(1+\frac{1}{{\mathrm{cot}}^{2}A}\right)=\frac{1}{{\mathrm{sin}}^{2}A-{\mathrm{sin}}^{4}A}$

Clara Reese

We use the basic trigonometry formula to solve the equation.
Prove: $\left(1+\frac{1}{{\mathrm{tan}}^{2}A}\right)\left(1+\frac{1}{{\mathrm{cot}}^{2}A}\right)=\frac{1}{{\mathrm{sin}}^{2}A-{\mathrm{sin}}^{4}A}$
$L.H.S.=\left(1+\frac{1}{{\mathrm{tan}}^{2}A}\right)\left(1+\frac{1}{{\mathrm{cot}}^{2}A}\right)$
$=\left(\frac{{\mathrm{tan}}^{2}A+1}{{\mathrm{tan}}^{2}A}\right)\left(\frac{{\mathrm{cot}}^{2}A+1}{{\mathrm{cot}}^{2}A}\right)$
$=\left(\frac{{\mathrm{sec}}^{2}A}{{\mathrm{tan}}^{2}A}\right)\left(\frac{{\mathrm{csc}}^{2}A}{{\mathrm{cot}}^{2}A}\right)$
$=\left(\frac{\frac{1}{{\mathrm{cos}}^{2}A}}{\frac{{\mathrm{sin}}^{2}A}{{\mathrm{cos}}^{2}A}}\right)\left(\frac{\frac{1}{{\mathrm{sin}}^{2}A}}{\frac{{\mathrm{cos}}^{2}A}{{\mathrm{sin}}^{2}A}}\right)$
$=\left(\frac{1}{{\mathrm{sin}}^{2}A}\right)\left(\frac{1}{{\mathrm{cos}}^{2}A}\right)$
$=\left(\frac{1}{{\mathrm{sin}}^{2}A}\right)\left(\frac{1}{1-{\mathrm{sin}}^{2}A}\right)$
$=\frac{1}{{\mathrm{sin}}^{2}A-{\mathrm{sin}}^{4}A}$
L.H.S=R.H.S
Hence prove.

Do you have a similar question?