2
B
7
COS
sin p =
s B =
tan B =
csc B =
sec B =
cot p =

Verified below

Explanation:

cot

B

sec

B

−

tan

B

−

cos

B

sec

B

+

tan

B

=

sin

B

+

csc

B

A common denominator to add these fractions:

cot

B

(

sec

B

+

tan

B

)

(

sec

B

−

tan

B

)

(

sec

B

+

tan

B

)

−

cos

B

(

sec

B

−

tan

B

)

(

sec

B

+

tan

B

)

(

sec

B

−

tan

B

)

=

sin

B

+

csc

B

cot

B

⋅

sec

B

+

cot

B

⋅

tan

B

sec

2

B

−

tan

2

B

−

cos

B

⋅

sec

B

−

cos

B

⋅

tan

B

sec

2

B

−

tan

2

B

Before we combine let's apply one Pythagorean identity:

1

+

tan

2

x

=

sec

2

x

Manipulated to:

1

=

sec

2

x

−

tan

2

x

That should look familiar as

sec

2

x

−

tan

2

x

is our denominator, so let's substitute a 1 in there:

cot

B

⋅

sec

B

+

cot

B

⋅

tan

B

1

−

cos

B

⋅

sec

B

−

cos

B

⋅

tan

B

1

There's still one more step before we combine, we need to utilize reciprocal and quotient identities to make the expression simpler:

tan

x

=

sin

x

cos

x

cot

x

=

1

tan

x

sec

x

=

1

cos

x

cot

x

=

cos

x

sin

x

1

sin

x

=

csc

x

So:

cos

B

sin

B

⋅

1

cos

B

+

1

tan

B

⋅

tan

B

1

−

cos

B

⋅

1

cos

B

−

cos

B

⋅

sin

B

cos

B

1

Leaves us with:

csc

B

+

1

−

(

1

−

sin

B

)

Distribute the negative:

csc

B

+

1

−

1

+

sin

B

Gets us to:

sin

B

+

csc

B

the correct answer is x/3+10≥32

hope this !

the answer is a if my teacher taught me correct