Question : If $\tan ^2 \theta+\tan ^4 \theta=1$, then:
Option 1: $\cot ^2 \theta+\cot ^4 \theta=1$
Option 2: $\cos ^2 \theta+\cos ^4 \theta=1$
Option 3: $\sin ^2 \theta+\sin ^4 \theta=1$
Option 4: $\operatorname{cosec}^2 \theta+\sec ^4 \theta=1$
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Correct Answer: $\cos ^2 \theta+\cos ^4 \theta=1$
Solution : $\tan ^2 \theta+\tan ^4 \theta=1$ ⇒ $\tan^2\theta(1+\tan^2\theta) = 1$ ⇒ $\tan^2\theta(1+\frac{\sin^2\theta}{\cos^2\theta}) = 1$ ⇒ $\frac{\sin^2\theta}{\cos^2\theta}(\frac{cos^2\theta + \sin^2\theta}{\cos^2\theta}) = 1$ ⇒ $\frac{\sin^2\theta}{\cos^2\theta}(\frac{1}{\cos^2\theta}) = 1$ ⇒ $\sin^2\theta = \cos^4\theta$ ⇒ $1-\cos^2\theta = \cos^4\theta$ ⇒ $\cos^4\theta + \cos^2\theta = 1$ Hence, the correct answer is $\cos^2\theta + \cos^4\theta = 1$.
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Question : $\left(\frac{\tan ^3 \theta}{\sec ^2 \theta}+\frac{\cot ^3 \theta}{\operatorname{cosec}^2 \theta}+2 \sin \theta \cos \theta\right) \div\left(1+\operatorname{cosec}^2 \theta+\tan ^2 \theta\right), 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : $\frac{(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)}{(\sec \theta+\tan \theta)(1-\sin \theta)}$ is equal to:
Question : Which of the following is equal to $[\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}]$?
Question : The value of $\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$ is:
Question : Let $0^{\circ}<\theta<90^{\circ}$, $\left(1+\cot ^2 \theta\right)\left(1+\tan ^2 \theta\right) × (\sin \theta-\operatorname{cosec} \theta)(\cos \theta-\sec \theta)$ is equal to:
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