Question

PLS HELP
question in image

Answer 1

Answer:

A = 9

B = 4 (If not included the minus)

Step-by-step explanation:

We are given the expression:

$\displaystyle \large{4 \tan (x) \sin (x) + 5 \sec(x) }$

We are going to rewrite the expression as Asec(x)-Bcos(x).

Let's recall the Identity of Trigonometric Function:

$\displaystyle \large{ \tan(x) = \frac{ \sin(x) }{ \cos(x) } } \\ \displaystyle \large{ \sec(x) = \frac{1}{ \cos(x) } } \\ \displaystyle \large{ { \sin}^{2} x = 1 - { \cos}^{2} x}$

We are going to use these identities to rewrite the expression. First, let's convert tan(x) to sin(x)/cos(x) and sec(x) to 1/cos(x).

$\displaystyle \large{4 ( \frac{ \sin(x) }{ \cos(x) } )\sin (x) + 5( \frac{1}{ \cos(x) }) }$

Simplify the expression — multiply sin(x) to sin(x)/cos(x).

$\displaystyle \large{4 ( \frac{ { \sin}^{2} x}{ \cos(x) } )+ 5( \frac{1}{ \cos(x) }) }$

Rewrite sin^2 of x as 1-cos^2(x).

$\displaystyle \large{4 ( \frac{ 1 - { \cos}^{2} x}{ \cos(x) } )+ 5( \frac{1}{ \cos(x) }) }$

Separate the fraction:

$\displaystyle \large{4 ( \frac{1}{ \cos(x) } - \frac{ { \cos }^{2} x}{ \cos(x) } )+ 5( \frac{1}{ \cos(x) }) }$

Rewrite 1/cos(x) as sec(x) again as we need it for rewrite.

$\displaystyle \large{4 ( \sec(x) - \frac{ { \cos }^{2} x}{ \cos(x) } )+ 5\sec(x) }$

Simplify cos^2 of x and cos(x).

$\displaystyle \large{4 ( \sec(x) - \frac{ \cos(x) }{ 1} )+ 5\sec(x) } \\ \displaystyle \large{4 ( \sec(x) - \cos(x) )+ 5\sec(x) }$

Expand 4 in sec(x)-cos(x).

$\displaystyle \large{4 \sec(x) - 4 \cos(x) + 5\sec(x) }$

Add 4sec(x) and 5sec(x).

$\displaystyle \large{9 \sec(x) - 4 \cos(x) }$

Hence, rewritten in form of Asec(x)-Bcos(x)

The value of A is 9 and B is 4 (If not included the sign).