Question

If $\frac{{\cos }^{4}x}{{\theta }_1}+\frac{{\sin }^{4}x}{{\theta }_2}=\frac{1}{{\theta }_1+{\theta }_2},$ then $\frac{{\theta }_2}{{\theta }_1}$ equals

• A. ${\cos }^{2}x$
• B. ${\sin }^{2}x$
• C. ${\tan }^{2}x$
• D. None of these

Answer 1

The correct option is C ${\tan }^{2}x$

$\frac{{cos}^{4}x}{{\theta }_1}+\frac{{sin}^{4}x}{{\theta }_2}=\frac{1}{{\theta }_1+{\theta }_2}\Rightarrow ({\theta }_1+{\theta }_2)(\frac{{cos}^{4}x}{{\theta }_1}+\frac{{sin}^{4}x}{{\theta }_2})=1\Rightarrow ({\theta }_1+{\theta }_2)(\frac{{cos}^{4}x}{{\theta }_1}+\frac{{sin}^{4}x}{{\theta }_2})={({sin}^{2}x+{cos}^{2}x)}^2\Rightarrow ({\theta }_1+{\theta }_2)(\frac{{cos}^{4}x}{{\theta }_1}+\frac{{sin}^{4}x}{{\theta }_2})={sin}^{4}x+{cos}^{4}x+2{sin}^{2}x{cos}^{2}x\Rightarrow ({\theta }_1+{\theta }_2){\theta }_2{cos}^{4}x+({\theta }_1+{\theta }_2){\theta }_1{sin}^{4}x={{\theta }_1{\theta }_{2}sin}^{4}x+{{\theta }_1{\theta }_{2}cos}^{4}x+{\theta }_1{\theta }_{2}2{sin}^{2}x{cos}^{2}x\Rightarrow {{\theta }_2}^2{cos}^{4}x+{{\theta }_1}^2{sin}^{4}x-{\theta }_1{\theta }_{2}2{sin}^{2}x{cos}^{2}x=0\Rightarrow \frac{{\theta }_2}{{\theta }_1}{cos}^{4}x+\frac{{\theta }_1}{{\theta }_2}{sin}^{4}x-2{sin}^{2}x{cos}^{2}x=0\Rightarrow {(\sqrt{\frac{{\theta }_2}{{\theta }_1}{cos}^{2}x}-\sqrt{\frac{{\theta }_1}{{\theta }_2}{sin}^{2}x})}^2=0\Rightarrow {tan}^{2}x=\frac{{\theta }_2}{{\theta }_1}$