Math, asked by vikasrahul335, 1 year ago

Prove that tan inverse of tan 2a + tan inverse cot a + tan inverse cos cube equals to zero

Answers

Answered by Anonymous
0

tan−1(2)+tan−1(3)=tan−1(

2+3

1−2⋅3

)=tan−1(−1)=nπ−

π

4

,

where n is any integer.

Now the principal value of tan−1(x) lies in [−

π

2

,

π

2

] precisely in (0,

π

2

) if finite x>0. So, the principal value of tan−1(2)+tan−1(3) will lie in (0,π).

So, the principal value of tan−1(2)+tan−1(3) will be

4

.

Interestingly, the principal value of tan−1(−1) is −

π

4

.

But the general values of tan−1(2)+tan−1(3) and tan−1(−1) are same.

Alternatively,

tan−1(1)+tan−1(2)+tan−1(3)=tan−1(

1+2+3−1⋅2⋅3

1−1⋅2−2⋅3−3⋅1

)=tan−1(0)=mπ

, where m is any integer.

Now the principal value of tan−1(1)+tan−1(2)+tan−1(3) will lie in (0,

2

) which is π.

The principal value of tan−1(0) is 0≠π.

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