Question : If $\sec 3 {x}=\operatorname{cosec}\left(3 {x}-45^{\circ}\right)$, where $3x$ is an acute angle, then ${x}$ is equal to:
Option 1: $45^{\circ}$
Option 2: $22.5^{\circ}$
Option 3: $35^{\circ}$
Option 4: $27.5^{\circ}$
Correct Answer: $22.5^{\circ}$
Solution : Given that $\sec 3x = \operatorname{cosec}(3x - 45^\circ)$, $⇒\frac{1}{\cos 3x} = \frac{1}{\sin(3x - 45^\circ)}$ $⇒\cos 3x = \sin(3x - 45^\circ)$ We know that $\sin(90^\circ - \theta) = \cos \theta$. $⇒ 90^\circ -3x = 3x - 45^\circ$ Solving for $x$, we get: $⇒6x = 135^\circ$ $⇒x = 22.5^\circ$ Hence, the correct answer is $22.5^\circ$.
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Question : If $\sin 3x=\cos(3x-45^{\circ}), 0^{\circ}<3x<90^{\circ}$, then $x$ is equal to:
Option 1: $45^\circ$
Option 2: $22.5^\circ$
Option 3: $35^\circ$
Option 4: $27.5^\circ$
Question : The value of $\tan ^2 48^{\circ}-\operatorname{cosec}^2 42^{\circ}+\operatorname{cosec}\left(67^{\circ}+\theta\right)-\sec \left(23^{\circ}-\theta\right)$ is:
Option 1: $-1$
Option 2: $0$
Option 3: $1$
Option 4: $-2$
Question : The value of $\frac{\sin ^2 30^{\circ}+\cos ^2 60^{\circ}-\sec 35^{\circ} \cdot \sin 55^{\circ}}{\sec 60^{\circ}+\operatorname{cosec} 30^{\circ}}$ is equal to:
Option 1: $\frac{1}{8}$
Option 2: $-\frac{1}{4}$
Option 3: $\frac{1}{4}$
Option 4: $-\frac{1}{8}$
Question : If $\theta$ is an acute angle and $\sin \theta+\operatorname{cosec} \theta=2$, then the value of $\sin ^5 \theta+\operatorname{cosec}^5 \theta$ is:
Option 1: 10
Option 2: 2
Option 3: 4
Option 4: 5
Question : If $\frac{1}{\operatorname{cosec} \theta+1}+\frac{1}{\operatorname{cosec} \theta-1}=2 \sec \theta, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{\tan \theta+2 \sec \theta}{\operatorname{cosec} \theta}$ is:
Option 1: $\frac{4+\sqrt{2}}{2}$
Option 2: $\frac{2+\sqrt{3}}{2}$
Option 3: $\frac{4+\sqrt{3}}{2}$
Option 4: $\frac{2+\sqrt{2}}{2}$
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