Question : If $2 \cot heta = 3$, find the value of $\frac{\sqrt{13} \sin heta – 3 an heta}{3 an heta + \sqrt{13} \cos heta}$
Option 1: $\frac{1}{\sqrt{13}}$
Option 2: $\frac{2}{\sqrt{13}}$
Option 3: 0
Option 4: $\frac{2}{3}$

Question

Question : If $2 \cot heta = 3$, find the value of $\frac{\sqrt{13} \sin heta – 3 an heta}{3 an heta + \sqrt{13} \cos heta}$

Option 1: $\frac{1}{\sqrt{13}}$

Option 2: $\frac{2}{\sqrt{13}}$

Option 3: 0

Option 4: $\frac{2}{3}$

Team Careers360 · Accepted Answer

Correct Answer: 0

Solution :
$2 \cot heta = 3$
$⇒\cot heta=\frac{3}{2}$
$⇒\frac{ ext{Base}}{ ext{Perpendicular}} =\frac{3}{2}$
$⇒\frac{AB}{BC}= \frac{3}{2}$
Let $AB = 3k, BC = 2k$
From Pythagoras' Theorem,
$AC^{2} = AB^{2} + BC^{2}$
$⇒AC^{2} = (3k)^{2} + (2k)^{2}$
$⇒AC^{2} = 13k^2$
$⇒AC = \sqrt{13}k$
$\sin heta = \frac{ ext{Perpendicular}}{ ext{Hypotenuse}} = \frac{2k}{\sqrt{13}k}=\frac{2}{\sqrt{13}}$
$\cos heta = \frac{ ext{Base}}{ ext{Hypotenuse}} = \frac{3k}{\sqrt{13}k}= \frac{3}{\sqrt{13}}$
$ an heta = \frac{ ext{Perpendicular}}{ ext{Base}} = \frac{2k}{3k}=\frac{2}{3}$
Putting these values in the equation, we get,
$=\frac{\sqrt{13} \sin heta – 3 an heta}{3 an heta + \sqrt{13} \cos heta}$
$=\frac{\sqrt{13} imes \frac{2}{\sqrt{13}} - 3 imes \frac{2}{3}}{3 imes\frac{2}{3} + \sqrt{13} imes \frac{3}{\sqrt{13}}}$
$=0$
Hence, the correct answer is 0.

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Question : If 2cot⁡θ=3, find the value of 13sin⁡θ–3tan⁡θ3tan⁡θ+13cos⁡θOption 1: 113Option 2: 213Option 3: 0Option 4: 23

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Question : If $2 \cot θ = 3$ , find the value of $\frac{\sqrt{13} \sin θ - 3 \tan θ}{3 \tan θ + \sqrt{13} \cos θ}$
Option 1: $\frac{1}{\sqrt{13}}$
Option 2: $\frac{2}{\sqrt{13}}$
Option 3: 0
Option 4: $\frac{2}{3}$