7.8. Conversion of Inverse Trigonometric Functions Using Geometric Figure or Figures.

Express `\sin^{-1}(\frac\alpha\beta)` into others inverse trigonometric functions.

Solution: Here, `\angle ACB=\sin^{-1}(\frac\alpha\beta)`.

    `BC=\sqrt(\beta^2-\alpha^2)`.

`\sin^{-1}\left(\frac\alpha\beta\right)=\cos^{-1}\left(\frac{\sqrt{\beta^2-\alpha^2}}\beta\right)`.

`\sin^{-1}\left(\frac\alpha\beta\right)=\tan^{-1}\left(\frac\alpha\{\sqrt{\beta^2-\alpha^2}))`.

`\sin^{-1}\left(\frac\alpha\beta\right)=\cot^{-1}\left(\frac{\sqrt{\beta^2-\alpha^2}}\alpha\right)`.

`\sin^{-1}\left(\frac\alpha\beta\right)=\sec^{-1}\left(\frac\beta\sqrt{\beta^2-\alpha^2})`.

`\sin^{-1}\left(\frac\alpha\beta\right)=\cosec^{-1}\left(\frac\beta\alpha)`.  

Similarly,

Express `\cos^{-1}(\frac\alpha\beta)` into others inverse trigonometric functions.

Express `\tan^{-1}(\frac\alpha\beta)` into others inverse trigonometric functions.

Express `\cot^{-1}(\frac\alpha\beta)` into others inverse trigonometric functions.

Express `\sec^{-1}(\frac\alpha\beta)` into others inverse trigonometric functions.

Express `\cosec^{-1}(\frac\alpha\beta)` into others inverse trigonometric functions.

Example

1. Show that, `\sin cot^{-1}\tan\cos^{-1}\frac3\4=\frac3\4`.

Solution:

 So, `\sin cot^{-1}\tan\cos^{-1}\frac3\4`

=`\sin cot^{-1}\tan\tan^{-1}\frac\sqrt7\3`

=`\sin cot^{-1}\frac\sqrt7\3`

=`\sin sin^{-1}\frac3\4`

=`\frac3\4`

(Proved)

2. Show that, `\sin cot^{-1}\tan\cos^{-1}x=x`.

Try yourself.

3. Show that, `\cos tan^{-1}\cot\sin^{-1}x=x`.

Try yourself.

4. Show that, `\cot cos^{-1}\sin\tan^{-1}x=x`.

Try yourself.

5. Show that, `\sin cos^{-1}\tan\sec^{-1}x=\sqrt(2-x^2)`.

Try yourself.

6. Show that, `\sin cot^{-1}\tan^{-1}x=\sqrt(\frac(1+x^2)(2+x^2))`.

Try yourself.

7. Show that, `\sin cos^{-1}\tan\sec^{-1}(\frac(x)(y))=\frac\sqrt(2y^2-x^2)y`.

Try yourself.