Question

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Answer 1

Topic

• Reciprocal trigonometric ratio.

$\implies \csc \theta =\dfrac{1}{\sin \theta }$

$\implies \sec\theta=\dfrac{1}{\cos \theta}$

$\implies \boxed{\cot \theta =\dfrac{1}{\tan \theta}=\dfrac{\cos \theta}{\sin \theta}}$

Solution

In this problem, the given expression needs to only contain $\cot \theta=\dfrac{\cos \theta }{\sin \theta }$.

The following is the given expression.

$\dfrac{4\sin \theta -3\cos \theta }{2\sin \theta +6\cos \theta }$

$=\dfrac{4\sin \theta -3\cos \theta }{2\sin \theta +6\cos \theta }\times \dfrac{\frac{1}{\sin \theta }}{\frac{1}{\sin \theta}}$

$=\dfrac{4-3\ \boxed{\cot \theta}}{2+6\ \boxed{\cot \theta }}$

$=\dfrac{4-2}{2+4}$

$=\dfrac{1}{3}$

This is the required answer.

Answer 2

$\overline{\underline{\boxed{\sf GIVEN \darr}}}$

If 3 cot θ = 2, find the value of :

$\frak{ \frac{4 \:\sin \theta - 3 \: cos \theta}{2 \: sin \theta + 6 \: cos \theta} }$

$\overline{\underline{\boxed{\sf SOLUTION \darr}}}$

3 cot θ = 2 can be written as cot θ = 2/3 [reason : by cross multiplication]

Now, we know that according to the square relations :

• cosec² θ - cot² θ = 1

Putting the value of cot θ we get

• cosec² θ - (2/3)² = 1

• cosec² θ - 4/9 = 1

Transposing - 4/9 to the other side we get + 4/9

• cosec² θ = 1 + 4/9

• cosec² θ = 13/9

• cosec θ = √13/3

Now, we know that sin θ = 1/cosec θ

sin θ = 3/√13

Again, tan θ = 1/cot θ

tan θ = 3/2

Now using quotient relations we get

$\frak{tan \theta = \frac{sin \theta}{cos \theta} }$

Putting the values of tan θ and sin θ we get

$\frak{ \frac{3}{2} = \frac{ \frac{3}{ \sqrt{13} } }{cos \theta} }$

By cross multiplication we see

$\frak{cos \theta} = \frac{ \cancel 3}{ \sqrt{13} } \times \frac{2}{ \cancel{3}}$

Therefore, cos θ = 2/√13

Now, according to the given condition which is :

$\frak{ \frac{4 \:\sin \theta - 3 \: cos \theta}{2 \: sin \theta + 6 \: cos \theta} }$

Putting the values we get

$\frak{ \frac{4 \times \frac{3}{ \sqrt{13} } - 3 \times \frac{2}{ \sqrt{13} } }{2 \times \frac{3}{ \sqrt{13}} + 6 \times \frac{2}{ \sqrt{13} } } }$

According to the BODMAS RULE multiplication is followed first then addition is done

So, multiplication here then addition

$\frak{ \frac{ \frac{12}{ \sqrt{13} } - \frac{6}{ \sqrt{13} } }{ \frac{6}{ \sqrt{13} } + \frac{12}{ \sqrt{13} } } }$

$\frak{ \frac{ \frac{6}{2 \sqrt{13} } }{ \frac{18}{2 \sqrt{13} } } }$

$\frak{ \frac{6}{ \cancel{2 \sqrt{13}} } \times \frac{ \cancel{2 \sqrt{13} }}{18} }$

$\frak{ \frac{6}{18} = \frac{1}{3} }$

So 1/3 is the answer