Question Number 189026 by Rupesh123 last updated on 10/Mar/23
Answered by Ar Brandon last updated on 10/Mar/23
$$\mathrm{cot}^{\mathrm{2}} {x}+\mathrm{1}=\mathrm{cosec}^{\mathrm{2}} {x}\: \\ $$$$\mid\mathrm{4}−{a}\mid+\mathrm{1}=\frac{\mid{a}\mid}{\mathrm{3}}\:,\:\mathrm{0}<{a}\leqslant\mathrm{4} \\ $$$$\Rightarrow\mathrm{4}−{a}+\mathrm{1}=\frac{{a}}{\mathrm{3}}\:\Rightarrow{a}=\frac{\mathrm{15}}{\mathrm{4}} \\ $$
Commented by Rupesh123 last updated on 10/Mar/23
Excellent!
Answered by manxsol last updated on 10/Mar/23
$${sinxctx}={cosx}=\sqrt{\frac{\mathrm{3}}{{a}}}\sqrt{\mathrm{4}−{a}} \\ $$$$\frac{\mathrm{3}}{{a}}+\frac{\mathrm{3}}{{a}}\left(\mathrm{4}−{a}\right)=\mathrm{1} \\ $$$$\frac{{a}}{\mathrm{3}}=\mathrm{5}−{a} \\ $$$${a}=\mathrm{15}−\mathrm{3}{a} \\ $$$$\mathrm{4}{a}=\mathrm{15} \\ $$$${a}=\mathrm{15}/\mathrm{4} \\ $$$$ \\ $$
Commented by Rupesh123 last updated on 10/Mar/23
Excellent
Answered by MJS_new last updated on 11/Mar/23
$$\mathrm{sin}^{\mathrm{2}} \:{x}\:=\frac{\mathrm{3}}{{a}}\:\Rightarrow\:\mathrm{cos}^{\mathrm{2}} \:{x}\:=\frac{{a}−\mathrm{3}}{{a}} \\ $$$$\Rightarrow\:\mathrm{cot}^{\mathrm{2}} \:{x}\:=\frac{{a}−\mathrm{3}}{\mathrm{3}}=\mathrm{4}−{a}\:\Rightarrow\:{a}=\frac{\mathrm{15}}{\mathrm{4}} \\ $$
Commented by manxsol last updated on 11/Mar/23
$${bien} \\ $$