248. From the relations established in the previous chapters, we may deduce corresponding relations connecting the inverse functions. Thus in the identity

and

3 cos-1α=cos-1 (4a3 — 3a).

249. To prove

that

tan-1x+tan-1y=tan -1

Let

x+y 1-xy'

tan-1x=a, so that tan a=x;
tan-1y=ß, so that tan ẞ=y.

We require a+ß in the form of an inverse tangent.

NOTE. The value of n cannot be assigned until we have selected

some particular values for the angles tan-15, tan-13, tan-1. choose the principal values, then n=0.

If we

16

=

--

sin-1

13

2

=cos-1

65

65*

We have to express a+ẞ as an inverse cosine.

Now cos (a+B) = cos a cos ẞ- sin a sin ß; whence by reading off the values of the functions from the figures in the margin, we have

It is sometimes convenient to work entirely in terms of the tangent or cotangent.

10. tan-1 +tan-1 +tan-1 =cot-13.

1

1

18. sin-1a-cos-1b-cos-1{b/1-a2+a√/1 − b2}.

24. tan (2 tan-1x)=2 tan (tan-1x+tan−1x3).

25. tan-1a-tan-1

α b 1+ab

+tan-
-1

b-c 1+be

+tan-1c.

26. If tan-1x+tan-1y+tan-1z=π, prove that

27. If u cot-1c cos a -tan-1/cos

x+y+z=xyz.

cos a, prove that

sin u=

_tan2 2.

250. We shall now shew how to solve equations expressed in the inverse notation.

also

tan (n+
(nx + 3T) =

4

-1;

Example 2. Solve sin-1+sin-1 (1-x)=cos-1x.

By transposition, sin-1 (1-x)=cos-1x-sin-1x.
Let cos-1x=a, and sin-1x=ẞ; then

But cos a=x, and therefore sin a= √1-x2; sin ẞ=x, and therefore cos ẞ=√1−x2;