Question : If $\operatorname{cosec} A+\cot A=3$, $0 \leq A \leq 90^{\circ}$, then find the value of cos A.
Option 1: $\frac{3}{4}$
Option 2: $\frac{2}{5}$
Option 3: $\frac{3}{5}$
Option 4: $\frac{4}{5}$
Correct Answer: $\frac{4}{5}$
Solution : Given, $\operatorname{cosec} A+\cot A=3$ ⇒ $\frac{1}{\sin A}+\frac{\cos A}{\sin A}=3$ ⇒ $1+\cos A = 3\sin A$ We know, $\sin^2A+\cos^2A=1$ ⇒ $1+\cos A = 3\sqrt{1-\cos^2A}$ Squaring both sides, ⇒ $(1+\cos A)^2=9(1-\cos^2A)$ ⇒ $1+\cos^2 A+2\cos A=9-9\cos^2A$ ⇒ $10\cos^2A+2\cos A-8=0$ ⇒ $5\cos^2A+\cos A - 4=0$ ⇒ $5\cos^2A+5\cos A-4\cos A - 4=0$ ⇒ $5\cos A(\cos A + 1)-4(\cos A + 1)=0$ ⇒ $(5\cos A-4)(\cos A + 1)=0$ ⇒ $\cos A = \frac45$ and $\cos A = -1$ $\cos A = -1$ will be rejected as $0 \leq A \leq 90^{\circ}$ ⇒ $\cos A = \frac45$ Hence, the correct answer is $\frac45$.
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Question : If $\tan A=\frac{4}{3}, 0 \leq A \leq 90^{\circ}$, then find the value of $\sin A$.
Option 1: $\frac{3}{5}$
Option 2: $1$
Option 3: $\frac{3}{4}$
Question : If $A+B=90^{\circ}$, then the expression $\frac{\cot A}{\cot B}+\cos ^2 A+\cos ^2 B$ is equal to:
Option 1: $\cot ^2 B$
Option 2: $\operatorname{cosec}^2 A$
Option 3: $\cot ^2 A$
Option 4: $\operatorname{cosec}^2 B$
Question : If $\cot \theta=\frac{1}{\sqrt{3}}, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{2-\sin ^2 \theta}{1-\cos ^2 \theta}+\left(\operatorname{cosec}^2 \theta-\sec \theta\right)$ is:
Option 1: 0
Option 2: 2
Option 3: 5
Option 4: 1
Question : If $\cos A=\frac{15}{17}, 0 \leq A \leq 90^{\circ}$, then the value of $\cot(90° - A)$ is:
Option 1: $\frac{8}{15}$
Option 2: $\frac{2 \sqrt{2}}{15}$
Option 3: $\frac{\sqrt{2}}{15}$
Option 4: $\frac{7}{15}$
Question : If $\sin A=\frac{\sqrt{3}}{2}, 0<A<90^{\circ}$, then find the value of $2(\operatorname{cosec} A + \cot A)$.
Option 1: $2 \sqrt{3}$
Option 2: $\sqrt{3}$
Option 3: $\frac{2}{\sqrt{3}}$
Option 4: $\frac{1}{\sqrt{3}}$
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