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Question Number 216421 by MATHEMATICSAM last updated on 07/Feb/25

If asinθ + bcosθ = acosecθ + bsecθ then prove that each term is equal to (a^(2/3) − b^(2/3) )(√(a^(2/3) + b^(2/3) )).

$$\mathrm{If}\:{a}\mathrm{sin}\theta\:+\:{b}\mathrm{cos}\theta\:=\:{a}\mathrm{cosec}\theta\:+\:{b}\mathrm{sec}\theta\:\mathrm{then} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{each}\:\mathrm{term}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left({a}^{\frac{\mathrm{2}}{\mathrm{3}}} \:−\:{b}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)\sqrt{{a}^{\frac{\mathrm{2}}{\mathrm{3}}} \:+\:{b}^{\frac{\mathrm{2}}{\mathrm{3}}} }. \\ $$

Answered by mr W last updated on 07/Feb/25

a sin θ+b cos θ=(a/(sin θ))+(b/(cos θ))=k, say a sin θ+b cos θ=k ...(i) a cos θ+b sin θ=k sin θ cos θ ...(ii) a cos θ+b sin θ=(a sin θ+b cos θ) sin θ cos θ a cos θ(1−sin^2 θ)+b sin θ (1−cos^2 θ)=0 a cos^3 θ+b sin^3 θ=0 ⇒tan θ=−(a^(1/3) /b^(1/3) ) ⇒sin θ=±(a^(1/3) /( (√(a^(2/3) +b^(2/3) )))) ⇒cos θ=∓(b^(1/3) /( (√(a^(2/3) +b^(2/3) )))) k=(a/(sin θ))+(b/(cos θ)) =±(a^(2/3) −b^(2/3) )(√(a^(2/3) +b^(2/3) )) ✓

$${a}\:\mathrm{sin}\:\theta+{b}\:\mathrm{cos}\:\theta=\frac{{a}}{\mathrm{sin}\:\theta}+\frac{{b}}{\mathrm{cos}\:\theta}={k},\:{say} \\ $$$${a}\:\mathrm{sin}\:\theta+{b}\:\mathrm{cos}\:\theta={k}\:\:\:...\left({i}\right) \\ $$$${a}\:\mathrm{cos}\:\theta+{b}\:\mathrm{sin}\:\theta={k}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\:\:...\left({ii}\right) \\ $$$${a}\:\mathrm{cos}\:\theta+{b}\:\mathrm{sin}\:\theta=\left({a}\:\mathrm{sin}\:\theta+{b}\:\mathrm{cos}\:\theta\right)\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta \\ $$$${a}\:\mathrm{cos}\:\theta\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\theta\right)+{b}\:\mathrm{sin}\:\theta\:\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\theta\right)=\mathrm{0} \\ $$$${a}\:\mathrm{cos}^{\mathrm{3}} \:\theta+{b}\:\mathrm{sin}^{\mathrm{3}} \:\theta=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=−\frac{{a}^{\frac{\mathrm{1}}{\mathrm{3}}} }{{b}^{\frac{\mathrm{1}}{\mathrm{3}}} } \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\pm\frac{{a}^{\frac{\mathrm{1}}{\mathrm{3}}} }{\:\sqrt{{a}^{\frac{\mathrm{2}}{\mathrm{3}}} +{b}^{\frac{\mathrm{2}}{\mathrm{3}}} }} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\mp\frac{{b}^{\frac{\mathrm{1}}{\mathrm{3}}} }{\:\sqrt{{a}^{\frac{\mathrm{2}}{\mathrm{3}}} +{b}^{\frac{\mathrm{2}}{\mathrm{3}}} }} \\ $$$${k}=\frac{{a}}{\mathrm{sin}\:\theta}+\frac{{b}}{\mathrm{cos}\:\theta} \\ $$$$\:\:\:=\pm\left({a}^{\frac{\mathrm{2}}{\mathrm{3}}} −{b}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)\sqrt{{a}^{\frac{\mathrm{2}}{\mathrm{3}}} +{b}^{\frac{\mathrm{2}}{\mathrm{3}}} }\:\:\:\checkmark \\ $$

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