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Question Number 131296 by EDWIN88 last updated on 03/Feb/21

If asin^(−1) (x)−bcos^(−1) (x)= c then the value of asin^(−1) (x)+bcos^(−1) (x) (whenever exists) is equal to ?

$$\:{If}\:{a}\mathrm{sin}^{−\mathrm{1}} \left({x}\right)−{b}\mathrm{cos}^{−\mathrm{1}} \left({x}\right)=\:{c}\: \\ $$$${then}\:{the}\:{value}\:{of}\:{a}\mathrm{sin}^{−\mathrm{1}} \left({x}\right)+{b}\mathrm{cos}^{−\mathrm{1}} \left({x}\right)\: \\ $$$$\left({whenever}\:{exists}\right)\:{is}\:{equal}\:{to}\:? \\ $$

Answered by liberty last updated on 03/Feb/21

We have sin^(−1) (x)+cos^(−1) (x)=(π/2) then bsin^(−1) (x)+bcos^(−1) (x)= ((bπ)/2)...(i) asin^(−1) (x)−bcos^(−1) (x)= c ...(ii) ∴ on adding (i) and (ii) we get sin^(−1) (x)= ((((bπ)/2) + c)/(a+b)) and cos^(−1) (x)=((((aπ)/2)−c)/(a+b)) Therefore asin^(−1) (x)+bcos^(−1) (x)= ((πab+c(a−b))/(a+b))

$$\mathrm{We}\:\mathrm{have}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)+\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{x}\right)=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{then}\:\mathrm{bsin}^{−\mathrm{1}} \left(\mathrm{x}\right)+\mathrm{bcos}^{−\mathrm{1}} \left(\mathrm{x}\right)=\:\frac{\mathrm{b}\pi}{\mathrm{2}}...\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{asin}^{−\mathrm{1}} \left(\mathrm{x}\right)−\mathrm{bcos}^{−\mathrm{1}} \left(\mathrm{x}\right)=\:\mathrm{c}\:...\left(\mathrm{ii}\right) \\ $$$$\:\therefore\:\mathrm{on}\:\mathrm{adding}\:\left(\mathrm{i}\right)\:\mathrm{and}\:\left(\mathrm{ii}\right)\:\mathrm{we}\:\mathrm{get}\: \\ $$$$\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)=\:\frac{\frac{\mathrm{b}\pi}{\mathrm{2}}\:+\:\mathrm{c}}{\mathrm{a}+\mathrm{b}}\:\mathrm{and}\:\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{x}\right)=\frac{\frac{\mathrm{a}\pi}{\mathrm{2}}−\mathrm{c}}{\mathrm{a}+\mathrm{b}} \\ $$$$\mathrm{Therefore}\:\mathrm{asin}^{−\mathrm{1}} \left(\mathrm{x}\right)+\mathrm{bcos}^{−\mathrm{1}} \left(\mathrm{x}\right)=\:\frac{\pi\mathrm{ab}+\mathrm{c}\left(\mathrm{a}−\mathrm{b}\right)}{\mathrm{a}+\mathrm{b}} \\ $$

Commented by EDWIN88 last updated on 03/Feb/21

sehr interessante Erklrung Sir

$$ \\ $$$$\mathrm{sehr}\:\mathrm{interessante}\:\mathrm{Erklrung}\:\mathrm{Sir}\: \\ $$

Answered by mr W last updated on 03/Feb/21

t=sin^(−1) x (π/2)−t=cos^(−1) x asin^(−1) (x)−bcos^(−1) (x)=c ⇒at−b((π/2)−t)=c ⇒(a+b)t=c+((bπ)/2) ⇒t=(1/(a+b))(c+((bπ)/2)) asin^(−1) (x)+bcos^(−1) (x) =at+b((π/2)−t) =(a−b)t+((bπ)/2) =((a−b)/(a+b))(c+((bπ)/2))+((bπ)/2) =(((a−b)c+abπ)/(a+b))

$${t}=\mathrm{sin}^{−\mathrm{1}} {x} \\ $$$$\frac{\pi}{\mathrm{2}}−{t}=\mathrm{cos}^{−\mathrm{1}} {x} \\ $$$${a}\mathrm{sin}^{−\mathrm{1}} \left({x}\right)−{b}\mathrm{cos}^{−\mathrm{1}} \left({x}\right)={c} \\ $$$$\Rightarrow{at}−{b}\left(\frac{\pi}{\mathrm{2}}−{t}\right)={c} \\ $$$$\Rightarrow\left({a}+{b}\right){t}={c}+\frac{{b}\pi}{\mathrm{2}} \\ $$$$\Rightarrow{t}=\frac{\mathrm{1}}{{a}+{b}}\left({c}+\frac{{b}\pi}{\mathrm{2}}\right) \\ $$$${a}\mathrm{sin}^{−\mathrm{1}} \left({x}\right)+{b}\mathrm{cos}^{−\mathrm{1}} \left({x}\right)\: \\ $$$$={at}+{b}\left(\frac{\pi}{\mathrm{2}}−{t}\right) \\ $$$$=\left({a}−{b}\right){t}+\frac{{b}\pi}{\mathrm{2}} \\ $$$$=\frac{{a}−{b}}{{a}+{b}}\left({c}+\frac{{b}\pi}{\mathrm{2}}\right)+\frac{{b}\pi}{\mathrm{2}} \\ $$$$=\frac{\left({a}−{b}\right){c}+{ab}\pi}{{a}+{b}} \\ $$

Commented by EDWIN88 last updated on 03/Feb/21

nice

$${nice} \\ $$

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