7. If [tex]$4 \tan \theta = 3$[/tex], then show that [tex]\frac{4 \sin \theta + 3 \cos \theta}{8 \sin \theta + 5 \cos \theta} = \frac{6}{11}[/tex].

8. If [tex]\cos \theta = \frac{\sqrt{3}}{2}[/tex], show that [tex]4 \cos ^3 \theta - 3 \cos \theta = 0[/tex].



Answer :

Sure, let's solve these problems step by step.

### Problem 7

Given the equation [tex]\( 4 \tan \theta = 3 \)[/tex], we need to show that:

[tex]\[ \frac{4 \sin \theta + 3 \cos \theta}{8 \sin \theta + 5 \cos \theta} = \frac{6}{11} \][/tex]

First, solve for [tex]\(\tan \theta\)[/tex]:

[tex]\[ \tan \theta = \frac{3}{4} \][/tex]

Now, recall that:

[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]

From [tex]\(\tan \theta = \frac{3}{4}\)[/tex], we can assume:

[tex]\[ \sin \theta = 3k \quad \text{and} \quad \cos \theta = 4k \quad \text{for some constant } k \][/tex]

Now, substitute [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex] into the expression:

[tex]\[ \frac{4 \sin \theta + 3 \cos \theta}{8 \sin \theta + 5 \cos \theta} = \frac{4(3k) + 3(4k)}{8(3k) + 5(4k)} \][/tex]

Simplify the numerator and denominator:

[tex]\[ = \frac{12k + 12k}{24k + 20k} = \frac{24k}{44k} \][/tex]

The [tex]\(k\)[/tex] terms cancel out:

[tex]\[ = \frac{24}{44} = \frac{6}{11} \][/tex]

Thus, we have shown that:

[tex]\[ \frac{4 \sin \theta + 3 \cos \theta}{8 \sin \theta + 5 \cos \theta} = \frac{6}{11} \][/tex]

### Problem 8

Given [tex]\( \cos \theta = \frac{\sqrt{3}}{2} \)[/tex], we need to show that:

[tex]\[ 4 \cos^3 \theta - 3 \cos \theta = 0 \][/tex]

First, calculate [tex]\( \cos^3 \theta \)[/tex]:

[tex]\[ \cos^3 \theta = \left( \frac{\sqrt{3}}{2} \right)^3 = \frac{(\sqrt{3})^3}{2^3} = \frac{3\sqrt{3}}{8} \][/tex]

Now, substitute [tex]\(\cos \theta\)[/tex] and [tex]\(\cos^3 \theta\)[/tex] into the expression:

[tex]\[ 4 \cos^3 \theta - 3 \cos \theta = 4 \left( \frac{3\sqrt{3}}{8} \right) - 3 \left( \frac{\sqrt{3}}{2} \right) \][/tex]

Simplify the expression:

[tex]\[ = \frac{12\sqrt{3}}{8} - \frac{3\sqrt{3}}{2} \][/tex]

Convert the terms to have a common denominator:

[tex]\[ = \frac{12\sqrt{3}}{8} - \frac{12\sqrt{3}}{8} \][/tex]

Subtract the fractions:

[tex]\[ = \frac{12\sqrt{3} - 12\sqrt{3}}{8} = \frac{0}{8} = 0 \][/tex]

Thus, we have shown that:

[tex]\[ 4 \cos^3 \theta - 3 \cos \theta = 0 \][/tex]

These solutions verify the given mathematical statements.