Question

If 4 tan =3 evaluate 4sin - cos +1 4sin+cos-1

Answer 1

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Answer 2

Answer: $\frac{13}{11}$

Explanation:

We have,

$4 than \theta = 3$

$\implies tan \theta = \frac{3}{4}$

$\implies sin \theta = \frac{3}{\sqrt{ 3^2 + 4^2}}=\frac{3}{\sqrt{9+16}}=\frac{3}{sqrt{25}}=\frac{3}{5}$

Similarly,

$cos \theta = \frac{4}{5}$

Thus,

$\frac{4 sin \theta - cos \theta + 1}{4 sin \theta + cos \theta - 1}$

$=\frac{4\times \frac{3}{5}-\frac{4}{5}+1}{4\times \frac{3}{5}+ \frac{4}{5} - 1}$

$=\frac{\frac{12}{5}-\frac{4}{5}+1}{\frac{12}{5}+\frac{4}{5}-1}$

$=\frac{\frac{12-4+5}{5}}{\frac{12+4-5}{5}}$

$=\frac{13}{11}$

Hence,

$\frac{4 sin \theta - cos \theta + 1}{4 sin \theta + cos \theta - 1}=\frac{13}{11}$