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Question Number 210036 by peter frank last updated on 29/Jul/24

Answered by Prithwish last updated on 29/Jul/24

$$ab=\left(\frac{1-\cos {}^2\theta }{\cos \theta }\right)\left(\frac{1-\sin {}^2\theta }{\sin \theta }\right)$$$$ab=\sin \theta \cos \theta$$$$a^2+\stackrel{2}{b}=\sec {}^2\theta -2+\cos {}^2\theta +\mathrm{cosec}{}^2\theta -2+\sin {}^2\theta$$$$=\frac{\sin {}^2\theta +\cos {}^2\theta }{\sin {}^2\theta \cos {}^2\theta }-3$$$$(a^2+b^2+3)=\frac{1}{a^2{b}^2}$$$$a^2{b}^2(a^2+b^2+3)=1Proved$$

Commented by peter frank last updated on 29/Jul/24

$$\mathrm{thank}\mathrm{you}$$

Answered by Frix last updated on 29/Jul/24

$$t=\tan x$$$$a^2=\frac{t^4}{t^2+1}\wedge b^2=\frac{1}{t^2(t^2+1)}$$$$a^2{b}^2(a^2+b^2+3)=\frac{t^2}{{(t^2+1)}^2}\times \frac{{(t^2+1)}^2}{t^2}=1$$

Commented by peter frank last updated on 29/Jul/24

$$\mathrm{this}\mathrm{very}\mathrm{tough}\mathrm{solution}.\mathrm{can}\mathrm{explain}\mathrm{a}$$$$\mathrm{little}\mathrm{bit}$$

Commented by Frix last updated on 29/Jul/24

$$t=\tan x$$$$\Rightarrow$$$$\sin x=\frac{t}{\sqrt{t^2+1}}\cos x=\frac{1}{\sqrt{t^2+1}}$$$$\sec x=\frac{1}{\cos x}=\sqrt{t^2+1}\csc x=\frac{1}{\sin x}=\frac{\sqrt{t^2+1}}{t}$$$$\sin x-\csc x=\frac{t}{\sqrt{t^2+1}}-\frac{\sqrt{t^2+1}}{t}=-\frac{1}{t\sqrt{t^2+1}}$$$$\cos x-\sec x=\frac{1}{\sqrt{t^2+1}}-\sqrt{t^2+1}=-\frac{t^2}{\sqrt{t^2+1}}$$

Commented by peter frank last updated on 29/Jul/24

$$\mathrm{thank}\mathrm{you}\mathrm{for}\mathrm{your}\mathrm{clarification}$$

Answered by Spillover last updated on 29/Jul/24

Commented by klipto last updated on 29/Jul/24

$$\mathbf{bro}\mathbf{you}\mathbf{niggerian}?$$

Commented by peter frank last updated on 29/Jul/24

$$\mathrm{thank}\mathrm{you}$$

Answered by Spillover last updated on 29/Jul/24

Commented by peter frank last updated on 29/Jul/24

$$\mathrm{thank}\mathrm{you}$$

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