Question

if tanA=3/4 and A is not in the first quadrant then {sin(π/2+A) - cot(π-A)} / tan(3π/2-A) - cos(3π/2+A)​

Answer 1

Answer:

Given

π

<

x

<

3

π

2

and

tan

x

=

3

4

π

<

x

<

3

π

2

⇒

π

2

<

x

2

<

3

π

4

→

x

2

∈

2nd quadrant

This means

sin

(

x

2

)

→

+

v

e

cos

(

x

2

)

→

−

v

e

tan

(

x

2

)

→

−

v

e

Now

tan

x

=

3

4

⇒

2

tan

(

x

2

)

1

−

tan

2

(

x

2

)

=

3

4

⇒

8

tan

(

x

2

)

=

3

−

3

tan

2

(

x

2

)

⇒

3

tan

2

(

x

2

)

+

8

tan

(

x

2

)

−

3

=

0

⇒

3

tan

2

(

x

2

)

+

9

tan(x2)−tan(x2)−3=0

⇒

3tan(x2)(tan(x2)+3)−1(tan(x2)+3)=0

⇒(3tan

(x2)−1)(tan(x2)

+3)=0

This means

tan

(x2)

=13

→

not acceptable as

tan(x2)

→−ve

So

tan(x2)

→−3

Now

cos(x2)

=1sec(x2)

=−1√1+tan2(x2)

=−1√1+(−3)2

=−1√10

Again

sin(x2)

=tan(x2)×cos(x2)

=−3×(−1√10)

=3√10