Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 169771 by greougoury555 last updated on 08/May/22

      (1/(1+cos^2 x)) + (1/(sin^2 x+1)) = ((48)/(35))

$$\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} {x}}\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{1}}\:=\:\frac{\mathrm{48}}{\mathrm{35}} \\ $$

Answered by kapoorshah last updated on 08/May/22

SyberMath

$${SyberMath} \\ $$

Answered by thfchristopher last updated on 08/May/22

⇒((1+sin^2 x+1+cos^2 x)/(1+sin^2 x+sin^2 xcos^2 x+cos^2 x))=((48)/(35))  ⇒(3/(2+sin^2 xcos^2 x))=((48)/(35))  ⇒105=96+48sin^2 xcos^2 x  ⇒sin^2 xcos^2 x=(9/(48))  ⇒(1/4)sin^2 2x=(9/(48))  ⇒sin^2 2x=((36)/(48))  ⇒sin 2x=((√3)/2)     or    sin 2x=−((√3)/2)  2x=nπ∓(π/3)  x=((nπ)/2)∓(π/6)

$$\Rightarrow\frac{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{1}+\mathrm{cos}^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{sin}^{\mathrm{2}} {x}\mathrm{cos}^{\mathrm{2}} {x}+\mathrm{cos}^{\mathrm{2}} {x}}=\frac{\mathrm{48}}{\mathrm{35}} \\ $$$$\Rightarrow\frac{\mathrm{3}}{\mathrm{2}+\mathrm{sin}^{\mathrm{2}} {x}\mathrm{cos}^{\mathrm{2}} {x}}=\frac{\mathrm{48}}{\mathrm{35}} \\ $$$$\Rightarrow\mathrm{105}=\mathrm{96}+\mathrm{48sin}^{\mathrm{2}} {x}\mathrm{cos}^{\mathrm{2}} {x} \\ $$$$\Rightarrow\mathrm{sin}^{\mathrm{2}} {x}\mathrm{cos}^{\mathrm{2}} {x}=\frac{\mathrm{9}}{\mathrm{48}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}^{\mathrm{2}} \mathrm{2}{x}=\frac{\mathrm{9}}{\mathrm{48}} \\ $$$$\Rightarrow\mathrm{sin}^{\mathrm{2}} \mathrm{2}{x}=\frac{\mathrm{36}}{\mathrm{48}} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{2}{x}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\:\:\:\mathrm{or}\:\:\:\:\mathrm{sin}\:\mathrm{2}{x}=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{2}{x}={n}\pi\mp\frac{\pi}{\mathrm{3}} \\ $$$${x}=\frac{{n}\pi}{\mathrm{2}}\mp\frac{\pi}{\mathrm{6}} \\ $$

Answered by cortano1 last updated on 08/May/22

Terms of Service

Privacy Policy

Contact: info@tinkutara.com