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Question Number 56525 by Sr@2004 last updated on 18/Mar/19

Commented by Sr@2004 last updated on 18/Mar/19

$$pleasesolve13and15$$

Commented by Sr@2004 last updated on 18/Mar/19

$$sorry13and14$$

Answered by MJS last updated on 18/Mar/19

$$13$$$$\left(1\right)\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1\Rightarrow x=\frac{a(b-y\sin \theta )}{b\cos \theta }$$$$\left(2\right)\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2-b^2\Rightarrow x=\frac{by+(a^2-b^2)\sin \theta }{a\tan \theta }$$$$\frac{a(b-y\sin \theta )}{b\cos \theta }=\frac{by+(a^2-b^2)\sin \theta }{a\tan \theta }$$$$(a^2{\sin }^2\theta +b^2{\cos }^2\theta )y-a^{2}b{\sin }^3\theta -b^3\sin \theta {\cos }^2\theta =0$$$$(a^2{\sin }^2\theta +b^2{\cos }^2\theta )(y-b\sin \theta )=0$$$$\Rightarrow y=b\sin \theta \Rightarrow x=a\cos \theta \Rightarrow$$$$\Rightarrow \sin \theta =\frac{y}{b}\wedge \cos \theta =\frac{x}{a}$$$$\mathrm{insert}:$$$$\Rightarrow \left(1\right)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\wedge \left(2\right)\mathrm{true}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Mar/19

$$13)\frac{xcos\theta }{a}+\frac{ysin\theta }{b}=1$$$$xp+yq=1\times \frac{1}{q}[p=\frac{cos\theta }{a}q=\frac{sin\theta }{b}]$$$$\frac{x}{p}-\frac{y}{q}=a^2-b^2\times q$$$$\frac{xp}{q}+y=\frac{1}{q}$$$$\frac{xq}{p}-y=q(a^2-b^2)$$$$x\left(\frac{p^2+q^2}{pq}\right)=q(a^2-b^2)+\frac{1}{q}$$$$x=\frac{q^2(a^2-b^2)+1}{q}\times \frac{pq}{(p^2+q^2)}$$$$x=\frac{cos\theta }{a}\times \frac{1}{{\left(\frac{cos\theta }{a}\right)}^2+{\left(\frac{sin\theta }{b}\right)}^2}\times [{\left(\frac{sin\theta }{b}\right)}^2(a^2-b^2)+1]$$$$x=\frac{cos\theta }{a}\times \frac{a^2{b}^2}{b^2{cos}^2\theta +a^2{sin}^2\theta }\times \left[\frac{b^2+a^2{sin}^2\theta -b^2{sin}^2\theta }{b^2}\right]$$$$\frac{x}{a}=cos\theta \times \frac{1}{b^2{cos}^2\theta +a^2{sin}^2\theta }\times [a^2{sin}^2\theta +b^2{cos}^2\theta ]$$$$\frac{x}{a}=cos\theta$$$$\frac{xcos\theta }{a}+\frac{ysin\theta }{b}=1$$$$\frac{ysin\theta }{b}=1-{cos}^2\theta$$$$\frac{y}{b}=sin\theta$$$${\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2$$$${cos}^2\theta +{sin}^2\theta$$$$=1proved$$

Commented by Sr@2004 last updated on 18/Mar/19

$$Icannotunderstandit.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Mar/19

$$14)sec\theta =\frac{1+sin\alpha sin\beta }{cos\alpha cos\beta }$$$${tan}^2\theta ={\left(\frac{1+sin\alpha sin\beta }{cos\alpha cos\beta }\right)}^2-1$$$$=\frac{1+2sin\alpha sin\beta +{sin}^2\alpha {sin}^2\beta -{cos}^2\alpha {cos}^2\beta }{{cos}^2\alpha {cos}^2\beta }$$$$letsin\alpha =acos\alpha =b$$$$sin\beta =ccos\beta =d$$$$RHS$$$$=\frac{1+2ac+a^2{c}^2-b^2{d}^2}{b^2{d}^2}$$$$=\frac{1+2ac+a^2{c}^2-(1-a^2)(1-c^2)}{b^2{d}^2}$$$$=\frac{1+2ac+a^2{c}^2-1+c^2+a^2-a^2{c}^2}{b^2{d}^2}$$$$={\left(\frac{a+c}{bd}\right)}^2$$$$tan\theta =\pm \left(\frac{a+c}{bd}\right)$$$$=\pm (\frac{sin\alpha }{cos\alpha cos\beta }+\frac{sin\beta }{cos\alpha cos\beta })$$$$=\pm (tan\alpha sec\beta +tan\beta sec\alpha )proved$$

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Mar/19

$$13)let\frac{xcos\theta }{a}+\frac{ysin\theta }{b}=1$$$$bxcos\theta +aysin\theta =ab$$$$x\left(bcos\theta \right)+y\left(asin\theta \right)=ab....\left(1\right)$$$$\frac{ax}{cos\theta }-\frac{by}{sin\theta }=a^2-b^2$$$$x\left(asin\theta \right)+y(-bcos\theta )=(a^2-b^2)sin\theta cos\theta .\times asin\theta$$$$x\left(bcos\theta \right)+y\left(asin\theta \right)=ab\times bcos\theta$$$$$$$$x(a^2{sin}^2\theta +b^2{cos}^2\theta )=a(a^2-b^2){sin}^2\theta cos\theta +{ab}^{2}cos\theta$$$$x(a^2{sin}^2\theta +b^2{cos}^2\theta )=acos\theta [a^2{sin}^2\theta -b^2{sin}^2\theta +b^2]$$$$x(a^2{sin}^2\theta +b^2{cos}^2\theta )=acos\theta [a^2{sin}^2\theta +b^2{cos}^2\theta ]$$$$x=acos\theta$$$$\frac{xcos\theta }{a}+\frac{ysin\theta }{b}=1$$$${cos}^2\theta +\frac{ysin\theta }{b}={sin}^2\theta +{cos}^2\theta$$$$\frac{ysin\theta }{b}={sin}^2\theta$$$$y=bsin\theta$$$$so$$$${\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2$$$${cos}^2\theta +{sin}^2\theta =1proved$$$$\frac{}{}$$

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