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Question Number 63300 by Rio Michael last updated on 02/Jul/19

$$showthat$$$$a)1+tan(\frac{\pi }{4}+A)=\frac{2}{1-tanA}$$$$b)2cos2\theta sin\theta +9sin\theta +3\equiv 11sin\theta -4{sin}^3\theta +3$$

Commented by kaivan.ahmadi last updated on 02/Jul/19

$$a.1+\frac{tg\left(\frac{\pi }{4}\right)+tg\left(A\right)}{1-tg\left(\frac{\pi }{4}\right)tg\left(A\right)}=\frac{1-tg\left(A\right)+1+tg\left(A\right)}{1-tg\left(A\right)}=\frac{2}{1-tg\left(A\right)}$$$$b.2(1-2{sin}^2\theta )sin\theta +9sin\theta +3=$$$$2sin\theta -4{sin}^3\theta +9sin\theta +3=11sin\theta -4{sin}^3\theta +3$$

Commented by Rio Michael last updated on 02/Jul/19

$$thatsright$$

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