[最も欲しかった] Tan2 Theta

$\\Solution:\\\\ Considering\ RHS\\\\ =\tan^6A + 3\tan^2A.\sec^2A + 1\\\\ =\tan^6A + 3\tan^2A.(1+\tan^2A)+1\ \ \ \ \ \ [\because \sec^2A+\tan^2A=1]\\\\ =(\tan^2A)^3+ 1^3+3\tan^2A.(1+\tan^2A)\\\\ Above\ equation\ is\ of\ the\ form: a^3+b^3+3ab(a+b)=(a+b)^3\\\\ Here, a=\tan^2A\ \ \ \and\ \ \ \ b= 1\\\\ =(\tan^2A+1)^3\\\\ =(\sec^2A)^3\\\\ =\sec^6A.=LHS\ \ \ \ \ \ [\because (a^m)^n=a^{m\times n}]\\\\ Hence\ proved.$

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$\sin^4\theta +sin ^2 \theta=1 \\=>sin^4\theta=cos^2\theta \\=>tan^2 \theta =cosec^2\theta \\=>tan^2 \theta =1+cot^2\theta \\=>x=1+1/x \\=>x^2-x-1=0 \\=>x={+1 \pm \sqrt{5} \over 2}$

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$\\ \frac{a^2}{x^2}+\frac{b^2}{y^2}=\frac{1}{\sin^2 \Theta }+\frac{1}{\tan^2 \Theta }\\\\\\ =\frac{1}{\sin^2 \Theta }+\frac{1}{\frac{\sin^2\Theta}{\cos^2\Theta}}\\\\\\ =\frac{1}{\sin^2 \Theta}+\frac{\cos^2\Theta}{\sin^2 \Theta} =\frac{1+\cos^2\Theta}{\sin^2 \Theta} \quad \quad [\because1+\cos^2\Theta= \sin^2 \Theta ]\\\\\\ =\frac{\sin^2\Theta}{\sin^2 \Theta}=1$

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$\sec^2 \theta + \tan^2 \theta = I\\ \frac{1}{\cos^2 \theta } + \frac{\sin^2 \theta}{\cos^2 \theta }= I \\ \frac{1+ \sin^2 \theta }{\cos^2 \theta }=I \\ \frac{1+ \sin^2 \theta }{1-\sin^2 \theta }=I \\ 1+ \sin^2 \theta = I(1- \sin^2 \theta )\\ \sin^2 \theta(1+I)= I-1 \\ \sin^2 \theta= \frac{(I-1)}{ I+1} \\ \sin \theta=\sqrt{ \frac{I-1}{ I+1} }\\$

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$\\\Rightarrow \frac{(\tan\theta -1)(\tan^2\theta+1+\tan\theta)}{\tan\theta(\tan\theta -1a)}\\\\ \Rightarrow \tan\theta+1+\frac{1}{\tan\theta}\\\\ \Rightarrow 1+\frac{1+\tan^2\theta}{\tan\theta}\\\\ \Rightarrow 1+\sec^2\theta \times \frac{1}{\tan\theta}\\\\ \Rightarrow 1+\sec\theta.\csc\theta\\\\ =RHS$

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$\cot^2\theta-\tan^2\theta=\frac{\cos^4\theta-\sin^4\theta}{\cos^2\theta\sin^2\theta} =4\dfrac{\cos2\theta}{\sin^22\theta}=4\dfrac{\cot2\theta}{\sin2\theta} \\ \\ \text{So we have } 4\cot2\theta(\sin2\theta-1)=0 \\ \\ \text{Now if } \cot2\theta=0\implies2\theta=n\pi+\dfrac\pi2 \text{ where } n \text{ is any integer} \\ \\ \text{Otherwise, }\sin2\theta=1\implies2\theta=2m\pi+\dfrac\pi2 \text{ where } n \text{ is any integer} \\ \\ \text{ Clearly, the second case is a subset of the first}$

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$\cos{n\theta }=\cos^n\theta \left(1-\binom{n}{2}\tan^{2}\theta +\binom{n}{4}\tan^{4}\theta+............ \right)\\ \cos{n\theta }=\left(\cos^n\theta -\binom{n}{2}\sin^{2}\theta \cos^{n-2}\theta +\binom{n}{4}\sin^{4}\theta \cos^{n-4}\theta +............ \right)\\ put \ \sin^2\theta=(1-\cos^2\theta ) \\ \text{we will get coefficeint of } \cos^n \theta =\frac{1+\binom{n}{1}+\binom{n}{2}+.......\binom{n}{n}}{2}\\\left(\text{taking in consideration that}\binom{n}{r}=\binom{n}{n-r} \right)\\ =2^{n-1}$

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$\\ \sin 2x = \frac{2\tan x}{1+\tan^2x}=\frac{2\left ( \frac{-1}{3} \right )}{1+\left ( \frac{-1}{3} \right )^2}=\frac{\frac{-2}{3}}{1+\frac{1}{9}}=\frac{\frac{-2}{3}}{\frac{10}{9}}=\frac{-3}{5}\\ \\ \\ \cos 2x = \frac{1-\tan^2x}{1+\tan^2x}=\frac{1-\left ( \frac{-1}{3} \right )^2}{1+\left ( \frac{-1}{3} \right )^2}=\frac{1-\frac{1}{9}}{1+\frac{1}{9}}=\frac{\frac{8}{9}}{\frac{10}{9}}=\frac{4}{5}\\ \\\\ \tan 2x = \frac{2\tan x}{1-\tan^2x}=\frac{2\left ( \frac{-1}{3} \right )}{1-\left ( \frac{-1}{3} \right )^2}=\frac{\frac{-2}{3}}{1-\frac{1}{9}}=\frac{\frac{-2}{3}}{\frac{8}{9}}=\frac{-3}{4}\\ \\\\ \\ \\ \cos 2\theta = 1-2\sin^2\theta =1-\left ( \frac{13}{85} \right )^2=1-\frac{169}{7225}=\frac{7056}{7225}$

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