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Question Number 170336 by Mastermind last updated on 21/May/22

$$if{cos}^2\propto +3{cosec}^2\theta =7,findthe$$$$expressionof{tan}^2\theta ,Henceshowthat$$$$tan\theta =1.$$$$$$$$Mastermind$$

Commented by MATHSLORD22 last updated on 21/May/22

$${\cos }^2\theta \mathrm{or}{\cos }^2\alpha ??$$

Answered by thfchristopher last updated on 21/May/22

$${\cos }^2\theta +{3\csc }^2\theta =7$$$$\Rightarrow {\sin }^2\theta {\cos }^2\theta +3={7\sin }^2\theta$$$$\Rightarrow {\sin }^2\theta (1-{\sin }^2\theta )+3={7\sin }^2\theta$$$$\Rightarrow {\sin }^2\theta -{\sin }^4\theta +3={7\sin }^2\theta$$$$\Rightarrow {\sin }^4\theta +{6\sin }^2\theta -3=0$$$${\sin }^2\theta =\frac{-6\pm \sqrt{36+12}}{2}=\frac{-6\pm 4\sqrt{3}}{2}$$$$\therefore {\sin }^2\theta =2\sqrt{3}-3$$$$\Rightarrow {\cos }^2\theta =4-2\sqrt{3}$$$$\Rightarrow {\tan }^2\theta =\frac{2\sqrt{3}-3}{4-2\sqrt{3}}$$$$=\frac{\sqrt{3}}{2}$$

Commented by MATHSLORD22 last updated on 21/May/22

$$\mathrm{He}\mathrm{wrote}\alpha \mathrm{not}\theta$$

Commented by mr W last updated on 21/May/22

$$\mathrm{He}\mathrm{wrote}\propto \mathrm{not}\alpha :)$$

Commented by Mastermind last updated on 21/May/22

$$Thankyouman$$

Answered by 2kdw last updated on 28/Apr/23

$$i){cos}^2\alpha +\frac{3}{{sin}^2\alpha }=7$$$$$$$$if{cos}^{2}x=1-{sin}^{2}x...\left(1\right)$$$$and{sin}^{2}x=t...\left(2\right)$$$$$$$$\therefore t(1-t)+3=7t$$$$-t^2+t+3=7t$$$$-t^2-6t+3=0$$$$\therefore {(t+3)}^2=12$$$$\Rightarrow t=2\sqrt{3}-3ort=-3-2\sqrt{3}$$$$$$$$but\left(1\right)1-t⩾0thent⩽1$$$$\therefore \overline{)\begin{array}{c}t=2\sqrt{3}-3\end{array}}$$$$$$$$ii)If\left(2\right)t={sin}^2\alpha$$$$[{sin}^2\alpha =2\sqrt{3}-3]$$$$$$$$...\left(1\right){cos}^2\alpha =1-2\sqrt{3}+3$$$$$$$$$$$$\therefore {tan}^2\alpha =\frac{2\sqrt{3}-3}{1-2\sqrt{3}+3}$$$$$$$$\overline{)\begin{array}{c}{tan}^2\alpha =\frac{2\sqrt{3}-3}{4-2\sqrt{3}}\end{array}}$$$$$$$$$$$$$$$$$$

Commented by Mastermind last updated on 21/May/22

$$Thanksman$$

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