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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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$\sec^2 \theta = 1 + \tan^2 \theta \\ RHS = 1 + \tan^2 \theta \\\\ = 1 + \frac{\sin^2 \theta }{\cos ^2 \theta } \\\\ = \frac{\cos^2 \theta + \sin^2 \theta }{\cos^2 \theta } \because \sin^2 \theta + \cos^2 \theta =1 \\\\ = \frac{1}{\cos ^2 \theta } \\\\ = \sec^2 \theta = LHS \\ $ Hence proved$

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$\\Solution:\\\\ Given: 1+\sin^2\Theta =3\sin\Theta.cos\Theta\\\\ \Rightarrow \sin^2\Theta+\cos^2\Theta+\sin^2\Theta =3\sin\Theta.cos\Theta\ \ \ \ \ \ \ \ [\because \sin^2\Theta+\cos^2\Theta=1]\\\\ \Rightarrow \sin^2\Theta+\cos^2\Theta=3\sin\Theta.cos\Theta\\\\ \Rightarrow 2\sin^2\Theta-3\sin\Theta.\cos\Theta+\cos^2\Theta=0\\\\ \Rightarrow 2\sin^2\Theta-2\sin\Theta.\cos\Theta-\sin\Theta.cos\Theta+\cos^2\Theta=0\\\\ \Rightarrow 2\sin\Theta(\sin\Theta-cos\Theta)-cos\Theta(\sin\Theta-cos\Theta)=0\\\\ \Rightarrow (\sin\Theta-cos\Theta)(2\sin\Theta-cos\Theta)=0\\\\ \Rightarrow (\sin\Theta-cos\Theta)=0\ \ or\ \ \ (2\sin\Theta-cos\Theta)=0\\\\ \Rightarrow \sin\Theta=cos\Theta\ \ \ or\ \ \ 2\sin\Theta=cos\Theta\\\\ \Rightarrow \frac{\sin\Theta}{cos\Theta}=1\ \ \ \ \ \ or\ \ \ \ \ \ \frac{\sin\Theta}{cos\Theta}=\frac12\\\\\\ \therefore \tan\Theta=1\ \ \ \ \ \ or\ \ \ \ \ \ \tan\Theta=\frac12$

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

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Tan2 Theta

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