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Given $\tan^{-1}y = \tan^{-1}x + \tan^{-1}\!\left(\dfrac{2x}{1-x^2}\right)$ with $|x|<\dfrac{1}{\sqrt{3}}$, find $y$.
Use $\tan^{-1}\!\left(\dfrac{2x}{1-x^2}\right)=2\tan^{-1}x$ (valid for $|x|<1$).
So $\tan^{-1}y = \tan^{-1}x + 2\tan^{-1}x = 3\tan^{-1}x$.
Using the triple angle formula: $\tan(3\theta)=\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$:
$y = \tan(3\tan^{-1}x) = \dfrac{3x-x^3}{1-3x^2}$
A company's bank reconciliation shows:
| Cash book balance (31 Dec) | $2,075 debit |
| Bank statement balance (31 Dec) | $2,250 credit |
| Bank charges not in cash book | $150 |
| Outstanding cheques | $325 |
What bank balance should be reported in the financial statements?
The financial statements show the updated cash book balance. Updated Cash Book = Draft Cash Book ($2,075) - Bank Charges ($150) = $1,925. Note that outstanding cheques are used to reconcile the bank statement to the cash book, not the other way around.
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