Question

If 3sin–1(2x1+x2) −4cos–1(1–x21+x2)+2tan–1(2x1–x2)=π3. Then, x=.​

Answer 1

Answer:

3sin

−1

[

1+x

2

2x

]−4cos

−1

[

1+x

2

1−x

2

]+2tan

−1

[

1−x

2

2x

]=

3

π

2tan

−1

x=tan

−1

[

1−x

2

2x

]=sin

−1

[

1+x

2

2x

]=cos

−1

[

1+x

2

1−x

2

]

∴ 3[2tan

−1

x]−4[2tan

−1

x]+2[2tan

−1

x]=

3

π

⇒6tan

−1

x−8tan

−1

x+4tan

−1

x=

3

π

⇒2tan

−1

x=

3

π

⇒tan

−1

x=

6

π

⇒x=tan

6

π

⇒x=

3

1