Question

If 3 cot theta = 2, then the value of 4sintheta - 3costheta / 2sintheta + 6costheta is​

Answer 1

⚝ Given :-

• 3 cot θ = 2

⚝ Find :-

• 4 sinθ - 3 cosθ/ 2 sinθ + 6 cosθ

⚝ Solution :-

$\leadsto{ \sf{3 \: Cot \theta = 2}}$

$\leadsto{ \sf{ Cot \theta = \frac{2}{3}}}$

• As we know that,

$\leadsto{ \sf{cot \theta = \frac{B}{P} }}$

• we can say that –

$\leadsto{ \sf{\frac{B}{P} = \frac{2}{3} }}$

• Thus,

• Base (B) = 2
• Perpendicular (P) = 3

• CALCULATING HYPOTENUSE :

By Using Pythagoras Theorem :-

⇝ H² = B² + P²

⇝ H² = 2² + 3²

⇝ H² = 4 + 9

⇝ H² = 13

⇝ H = √13

• Therefore, Hypotenuse is √13. Now,

$\leadsto{ \boxed{ \sf{sin \theta = \frac{P}{H} }}}$

• Substitute the values :

$\leadsto{ \boxed{ \sf{sin \theta = \frac{3}{ \sqrt{13} } }}} \: \bigstar$

• Similarly,

$\leadsto{ \boxed{ \sf{cos \theta = \frac{B}{H} }}}$

• Substitute the values :

$\leadsto{ \boxed{ \sf{cos \theta = \frac{2}{ \sqrt{13} } }}} \bigstar$

Now,

$\leadsto{ \displaystyle{ \sf{ \frac{4 \: sin \theta - 3\: cos \theta}{2\: sin \theta + 6\: cos \theta}}}}$

Now, we've –

• Sin θ = 3/√13
• Cos θ = 2/√13

Putting all values :–

$\leadsto{ \displaystyle{ \sf{ \frac{4 \times \frac{3}{ \sqrt{13} } \: - \: 3 \times \frac{2}{ \sqrt{13} } }{2 \times \frac{3}{ \sqrt{13} } \: + \: 6 \times \frac{2}{ \sqrt{13} } }}}}$

$\leadsto{ \displaystyle{ \sf{ \frac{ \frac{12}{ \sqrt{13} } - \frac{6}{ \sqrt{13} } }{ \frac{6}{ \sqrt{13} } + \frac{12}{ \sqrt{13} } } }}}$

$\leadsto{ \displaystyle{ \sf{ \frac{ \frac{6}{ \sqrt{13} } }{ \frac{18}{ \sqrt{13} }}}}}$

$\leadsto {\displaystyle{ \sf{\frac{6}{ \sqrt{13} } \times \frac{ \sqrt{13} }{18} }}}$

$\leadsto {\displaystyle{ \sf{\frac{ \cancel6}{ \cancel{\sqrt{ 13}} } \times \frac{ \cancel{\sqrt{13}} }{ \cancel{18} \: \: ^{3} } }}}$

$\leadsto{ \displaystyle{ \sf{ \red{ \frac{1}{3} }}}}$

• Henceforth, the value of $\bold{ \sf{ \frac{4 \: sin \theta - 3\: cos \theta}{2\: sin \theta + 6\: cos \theta}}}$ is $\bold{ \sf{ \frac{1}{3}}}$

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