Question : Value of $\sec^2\theta-\frac{\sin^2\theta-2\sin^4\theta}{2\cos^4\theta-\cos^2\theta}$ is:
Option 1: 1
Option 2: 2
Option 3: –1
Option 4: 0
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Correct Answer: 1
Solution : $\sec^2\theta-\frac{\sin^2\theta-2\sin^4\theta}{2\cos^4\theta-\cos^2\theta}$ $=\sec^2\theta-\frac{\sin^2\theta(1-2\sin^2\theta)}{\cos^2\theta(2\cos^2\theta-1)}$ $=\sec^2\theta-\frac{\sin^2\theta(1-2\sin^2\theta)}{\cos^2\theta[2(1-\sin^2\theta)-1]}$ $=\sec^2\theta-\frac{\sin^2\theta(1-2\sin^2\theta)}{\cos^2\theta(1-2\sin^2\theta)}$ $=\sec^2\theta-\tan^2\theta$ $=1$ Hence, the correct answer is 1.
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Question : If $\tan \theta=\frac{4}{3}$, then the value of $\frac{3\sin \theta+ 2\cos \theta}{3\sin \theta – 2 \cos \theta}$ is:
Question : Which of the following is equal to $[\frac{\tan \theta+\sec \theta–1}{\tan \theta–\sec \theta+1}]$?
Question : If $x\sin^{3}\theta +y\cos^{3}\theta=\sin\theta\cos\theta$ and $x\sin\theta-y\cos\theta=0$, then the value of $\left ( x^{2}+y^{2} \right )$ equals:
Question : $\frac{\sin^4 \theta+\cos^4 \theta}{1-2 \sin^2 \theta \cos^2 \theta}=$____.
Question : $\frac{(1+\sec \theta \operatorname{cosec} \theta)^2(\sec \theta-\tan \theta)^2(1+\sin \theta)}{(\sin \theta+\sec \theta)^2+(\cos \theta+\operatorname{cosec} \theta)^2}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
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