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Question Number 210036 by peter frank last updated on 29/Jul/24

Answered by Prithwish last updated on 29/Jul/24

ab=(((1−cos^2 θ)/(cos θ)))(((1−sin^2 θ)/(sin θ))) ab=sin θcos θ a^2 +b^2 = sec^2 θ−2+cos^2 θ+cosec^2 θ−2+sin^2 θ = ((sin^2 θ+cos^2 θ)/(sin^2 θcos^2 θ)) −3 (a^2 +b^2 +3)= (1/(a^2 b^2 )) a^2 b^2 (a^2 +b^2 +3)=1 Proved

$${ab}=\left(\frac{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{cos}\:\theta}\right)\left(\frac{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{sin}\:\theta}\right) \\ $$$${ab}=\mathrm{sin}\:\theta\mathrm{cos}\:\theta \\ $$$${a}^{\mathrm{2}} +\overset{\mathrm{2}} {{b}}\:=\:\mathrm{sec}\:^{\mathrm{2}} \theta−\mathrm{2}+\mathrm{cos}\:^{\mathrm{2}} \theta+\mathrm{cosec}\:^{\mathrm{2}} \theta−\mathrm{2}+\mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:^{\mathrm{2}} \theta}\:−\mathrm{3} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{3}\right)=\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\: \\ $$$${a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{3}\right)=\mathrm{1}\:{Proved} \\ $$

Commented by peter frank last updated on 29/Jul/24

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Answered by Frix last updated on 29/Jul/24

t=tan x a^2 =(t^4 /(t^2 +1))∧b^2 =(1/(t^2 (t^2 +1))) a^2 b^2 (a^2 +b^2 +3)=(t^2 /((t^2 +1)^2 ))×(((t^2 +1)^2 )/t^2 )=1

$${t}=\mathrm{tan}\:{x} \\ $$$${a}^{\mathrm{2}} =\frac{{t}^{\mathrm{4}} }{{t}^{\mathrm{2}} +\mathrm{1}}\wedge{b}^{\mathrm{2}} =\frac{\mathrm{1}}{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$${a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{3}\right)=\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }×\frac{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{{t}^{\mathrm{2}} }=\mathrm{1} \\ $$

Commented by peter frank last updated on 29/Jul/24

this very tough solution.can explain a little bit

$$\mathrm{this}\:\mathrm{very}\:\mathrm{tough}\:\mathrm{solution}.\mathrm{can}\:\mathrm{explain}\:\mathrm{a}\: \\ $$$$\mathrm{little}\:\mathrm{bit} \\ $$

Commented by Frix last updated on 29/Jul/24

t=tan x ⇒ sin x =(t/( (√(t^2 +1)))) cos x =(1/( (√(t^2 +1)))) sec x =(1/(cos x))=(√(t^2 +1)) csc x =(1/(sin x))=((√(t^2 +1))/t) sin x −csc x =(t/( (√(t^2 +1))))−((√(t^2 +1))/t)=−(1/(t(√(t^2 +1)))) cos x −sec x =(1/( (√(t^2 +1))))−(√(t^2 +1))=−(t^2 /( (√(t^2 +1))))

$${t}=\mathrm{tan}\:{x} \\ $$$$\Rightarrow \\ $$$$\mathrm{sin}\:{x}\:=\frac{{t}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}\:\:\:\:\:\mathrm{cos}\:{x}\:=\frac{\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\mathrm{sec}\:{x}\:=\frac{\mathrm{1}}{\mathrm{cos}\:{x}}=\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\:\:\:\:\:\mathrm{csc}\:{x}\:=\frac{\mathrm{1}}{\mathrm{sin}\:{x}}=\frac{\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}{{t}} \\ $$$$\mathrm{sin}\:{x}\:−\mathrm{csc}\:{x}\:=\frac{{t}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}−\frac{\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}{{t}}=−\frac{\mathrm{1}}{{t}\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\mathrm{cos}\:{x}\:−\mathrm{sec}\:{x}\:=\frac{\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}−\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}=−\frac{{t}^{\mathrm{2}} }{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}} \\ $$

Commented by peter frank last updated on 29/Jul/24

thank you for your clarification

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{your}\:\mathrm{clarification} \\ $$

Answered by Spillover last updated on 29/Jul/24

Commented by klipto last updated on 29/Jul/24

bro you niggerian?

$$\boldsymbol{\mathrm{bro}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{niggerian}}? \\ $$

Commented by peter frank last updated on 29/Jul/24

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Answered by Spillover last updated on 29/Jul/24

Commented by peter frank last updated on 29/Jul/24

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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