Question Number 210036 by peter frank last updated on 29/Jul/24
Answered by Prithwish last updated on 29/Jul/24
$${ab}=\left(\frac{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{cos}\:\theta}\right)\left(\frac{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{sin}\:\theta}\right) \\ $$$${ab}=\mathrm{sin}\:\theta\mathrm{cos}\:\theta \\ $$$${a}^{\mathrm{2}} +\overset{\mathrm{2}} {{b}}\:=\:\mathrm{sec}\:^{\mathrm{2}} \theta−\mathrm{2}+\mathrm{cos}\:^{\mathrm{2}} \theta+\mathrm{cosec}\:^{\mathrm{2}} \theta−\mathrm{2}+\mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:^{\mathrm{2}} \theta}\:−\mathrm{3} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{3}\right)=\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\: \\ $$$${a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{3}\right)=\mathrm{1}\:{Proved} \\ $$
Commented by peter frank last updated on 29/Jul/24
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by Frix last updated on 29/Jul/24
$${t}=\mathrm{tan}\:{x} \\ $$$${a}^{\mathrm{2}} =\frac{{t}^{\mathrm{4}} }{{t}^{\mathrm{2}} +\mathrm{1}}\wedge{b}^{\mathrm{2}} =\frac{\mathrm{1}}{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$${a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{3}\right)=\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }×\frac{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{{t}^{\mathrm{2}} }=\mathrm{1} \\ $$
Commented by peter frank last updated on 29/Jul/24
$$\mathrm{this}\:\mathrm{very}\:\mathrm{tough}\:\mathrm{solution}.\mathrm{can}\:\mathrm{explain}\:\mathrm{a}\: \\ $$$$\mathrm{little}\:\mathrm{bit} \\ $$
Commented by Frix last updated on 29/Jul/24
$${t}=\mathrm{tan}\:{x} \\ $$$$\Rightarrow \\ $$$$\mathrm{sin}\:{x}\:=\frac{{t}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}\:\:\:\:\:\mathrm{cos}\:{x}\:=\frac{\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\mathrm{sec}\:{x}\:=\frac{\mathrm{1}}{\mathrm{cos}\:{x}}=\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\:\:\:\:\:\mathrm{csc}\:{x}\:=\frac{\mathrm{1}}{\mathrm{sin}\:{x}}=\frac{\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}{{t}} \\ $$$$\mathrm{sin}\:{x}\:−\mathrm{csc}\:{x}\:=\frac{{t}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}−\frac{\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}{{t}}=−\frac{\mathrm{1}}{{t}\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\mathrm{cos}\:{x}\:−\mathrm{sec}\:{x}\:=\frac{\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}−\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}=−\frac{{t}^{\mathrm{2}} }{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}} \\ $$
Commented by peter frank last updated on 29/Jul/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{your}\:\mathrm{clarification} \\ $$
Answered by Spillover last updated on 29/Jul/24
Commented by klipto last updated on 29/Jul/24
$$\boldsymbol{\mathrm{bro}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{niggerian}}? \\ $$
Commented by peter frank last updated on 29/Jul/24
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by Spillover last updated on 29/Jul/24
Commented by peter frank last updated on 29/Jul/24
$$\mathrm{thank}\:\mathrm{you} \\ $$