Question

(t/2-1/3) (t/2-1/3) and (a-2/3) (a-2/3)​

Answer 1

Answer:

We know that,

tan2x=

1−tan

2

x

2tanx

& tan3x=

1−3tan

2

x

3tanx−tan

3

x

So, we clearly have

f(tanx)=tan2x&g(tanx)=tan3x

let us simplify the value of t

t=

3

−1

2

2

−(1+

3

)

rationalizing the denominator,

t=

(

3

−1)(

3

+1)

(2

2

−(1+

3

))(

3

+1)

after simplification,

t=

2

−

3

−

4

+

6

⟹t=tan7.5 (here 7.5 is in degrees)

Now, to find

dt

d

{f(g(t))} let us assume

t=tanx (where x=7.5)

differentiating on both sides, we have

⟹dt=sec

2

xdx

⟹

dt

dx

=cos

2

x

So,

⟹

dt

d

{f(g(t))}=

dt

d

{f(g(tanx))}=

dt

d

{f(tan3x)}=

dt

d

{tan6x}

=(sec

2

6x)(6)(

dt

dx

)=(sec

2

6x)(6)(cos

2

x)

put x=7.5,

we get,

=(12)cos

2

7.5

The value becomes

(

2

12+3

6

+3

2

).