Prove that:

$(1+\frac{1}{{\mathrm{tan}}^{2}A})(1+\frac{1}{{\mathrm{cot}}^{2}A})=\frac{1}{{\mathrm{sin}}^{2}A-{\mathrm{sin}}^{4}A}$

amanf
2021-08-22
Answered

Prove that:

$(1+\frac{1}{{\mathrm{tan}}^{2}A})(1+\frac{1}{{\mathrm{cot}}^{2}A})=\frac{1}{{\mathrm{sin}}^{2}A-{\mathrm{sin}}^{4}A}$

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Clara Reese

Answered 2021-08-23
Author has **120** answers

We use the basic trigonometry formula to solve the equation.

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

$L.H.S.=(1+\frac{1}{{\mathrm{tan}}^{2}A})(1+\frac{1}{{\mathrm{cot}}^{2}A})$

$=\left(\frac{{\mathrm{tan}}^{2}A+1}{{\mathrm{tan}}^{2}A}\right)\left(\frac{{\mathrm{cot}}^{2}A+1}{{\mathrm{cot}}^{2}A}\right)$

$=\left(\frac{{\mathrm{sec}}^{2}A}{{\mathrm{tan}}^{2}A}\right)\left(\frac{{\mathrm{csc}}^{2}A}{{\mathrm{cot}}^{2}A}\right)$

$=\left(\frac{\frac{1}{{\mathrm{cos}}^{2}A}}{\frac{{\mathrm{sin}}^{2}A}{{\mathrm{cos}}^{2}A}}\right)\left(\frac{\frac{1}{{\mathrm{sin}}^{2}A}}{\frac{{\mathrm{cos}}^{2}A}{{\mathrm{sin}}^{2}A}}\right)$

$=\left(\frac{1}{{\mathrm{sin}}^{2}A}\right)\left(\frac{1}{{\mathrm{cos}}^{2}A}\right)$

$=\left(\frac{1}{{\mathrm{sin}}^{2}A}\right)\left(\frac{1}{1-{\mathrm{sin}}^{2}A}\right)$

$=\frac{1}{{\mathrm{sin}}^{2}A-{\mathrm{sin}}^{4}A}$

L.H.S=R.H.S

Hence prove.

Prove:

L.H.S=R.H.S

Hence prove.

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