Question

please answer the above ⬆️⬆️⬆️ question

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Answer 1

this question is of chapter:- trigonometry

sol:-

given:- 4tan0=3

tan0=3/4

AB/BC=3/4

let the ratios be k

AB=3k and BC=4k

In ∆ABC,

angle B= 90°

by Pythagoras theorem

AC²=(3k)²+(4k)²

AC²=9k²+16k²

AC²=25k²

AC=5k

sin0=AB/AC

=3k/5k

=3/5

cos0=BC/AC

=4k/5k

=4/5


$\frac{4 \sin(0) - \cos(0) + 1 }{4 \sin(0) + \cos(0) - 1 } = \frac{4 \times \frac{3}{5} - \frac{4}{5} + 1 }{4 \times \frac{3}{5} + \frac{4}{5} - 1}$

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

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

$\frac{4 \sin(0) - \cos(0) + 1 }{4 \sin(0) + \cos(0) - 1 } = \frac{13}{11}$

Answer 2

GOOD MORNING FRIENDS

ANSWER

given:- 4 tan0=3

tan0=3/4

AB/BC=3/4

let the ratios be k

AB=3k and BC=4k

In ∆ABC,

angle B= 90°

by Pythagoras theorem

AC²=(3k)²+(4k)²

AC²=9k²+16k²

AC²=25k²

AC=5k

sin

0=AB/AC

=3k/5k

=3/5

cos0=BC/AC

=4k/5k

=4/5

\frac{4 \sin(0) - \cos(0) + 1 }{4 \sin(0) + \cos(0) - 1 } = \frac{4 \times \frac{3}{5} - \frac{4}{5} + 1 }{4 \times \frac{3}{5} + \frac{4}{5} - 1}

4sin(0)+cos(0)−1

4sin(0)−cos(0)+1
=

4×

5

3

+

5

4

−1

4×

5

3

−

5

4

+1

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

5

12

+

5

4

−1

5

12

−

5

4

+1

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

12+4−5

12−4+5

\frac{4 \sin(0) - \cos(0) + 1 }{4 \sin(0) + \cos(0) - 1 } = \frac{13}{11}

4sin(0)+cos(0)−1

4sin(0)−cos(0)+1

=

11

13

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