Question : $\tan(θ -14\pi)$ is equal to:
Option 1: $\tan \theta$
Option 2: $-\cot \theta$
Option 3: $\cot \theta$
Option 4: $-\tan \theta$
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Correct Answer: $\tan \theta$
Solution : The function $\tan(θ)$ has a period of $\pi$. This means that for any integer $n$, the value of $\tan(n\pi-θ ) =-\tanθ$ $\tan(θ - 14\pi)=\tan(-(2\times 7\pi-\theta))$ $=-\tan(2\times 7\pi-\theta)=-\times (-\tanθ)$ $\therefore \tan(θ - 14\pi)=\tanθ$ Hence, the correct answer is $\tanθ$.
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Question : $(\sin \theta+\operatorname{cosec} \theta)^2+(\cos \theta+\sec \theta)^2=$?
Option 1: $5+\tan ^2 \theta+\cot ^2 \theta$
Option 2: $7+\tan ^2 \theta-\cot ^2 \theta$
Option 3: $7+\tan ^2 \theta+\cot ^2 \theta$
Option 4: $5+\tan ^2 \theta-\cot ^2 \theta$
Question : Which of the following is equal to $[\frac{\tan \theta+\sec \theta–1}{\tan \theta–\sec \theta+1}]$?
Option 1: $\frac{1+\sin \theta}{\cos \theta}$
Option 2: $\frac{1+\tan \theta}{\cot \theta}$
Option 3: $\frac{1+\cot \theta}{\tan \theta}$
Option 4: $\frac{1+\cos \theta}{\sin \theta}$
Question : The expression $\frac{(1-\sin \theta+\cos \theta)^2(1-\cos \theta) \sec ^3 \theta\; {\operatorname{cosec}}^2 \theta}{(\sec \theta-\tan \theta)(\tan \theta+\cot \theta)}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Option 1: $\sin \theta$
Option 2: $2 \cos \theta$
Option 4: $2 \tan \theta$
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: $\operatorname{cosec}^2 \theta+\sec ^4 \theta=1$
Question : $\frac{(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)}{(\sec \theta+\tan \theta)(1-\sin \theta)}$ is equal to:
Option 1: $2 \sec \theta$
Option 2: $2 \operatorname{cosec} \theta$
Option 3: $\operatorname{cosec} \theta$
Option 4: $\sec \theta$
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