=

(tanA+tanB)/( 1-tanAtanB)

And tan (quarter pie)

=

1tan (quarter +a)=2

= & gt

(Tan (Jipai)+

Tan (1) /( 1- Tan (a quarter of schools) Tan (1))

=

2

= & gt

( 1+tan(a))/( 1-tan(a))

=

2

= & gt 1+tan(a)=2-2tan(a)

= & gt

3tan(a)= 1

= & gt

Tan (1)

=

1/3 the first question and answer is

The value of tan(a) is 1/3. Question 2: (2sin? α+sin 2α)/( 1+tanα)=(2 sinαsinα+2 sinαcosα)/( 1+tanα)

=

2 sinαcosα(sinα/cosα+ 1)/( 1+tanα)

=

sin 2α(tanα+ 1)/( 1+tanα)

=

Sin2α can be obtained from the universal formula sinα = [2tan (α/2)]/{1+[tan (α/2)] 2} that sin2α = 2tan α/(1+tan α) = 2 * (1/3)/.

2/3/(10/9) = 6/10 = 3/5, so the value of the second question is 3/5.