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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