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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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