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- Thread starter MyHeadee
- Start date
Life'sHard
Well-Known Member
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- May 24, 2021
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- HSC
- 2021
- Uni Grad
- 2025
What are the previous parts to this question.
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Cheers, but that bit is done. iii) has tripped me}\\\text{Formula for} \\tan(\alpha -\beta) = \frac{\\tan(\alpha) - \\tan(\beta)}{1+\\tan(\alpha)\\tan(\beta)}\\\\\text{Therefore}\\\\\\tan(3\theta - 2\theta) = \frac{\\tan(3\theta) - \\tan(2\theta)}{1+\\tan(3\theta)\\tan(2\theta)}\\\\\\tan(\theta) = \frac{\\tan(3\theta) - \\tan(2\theta)}{1+\\tan(3\theta)\\tan(2\theta)}\\\\\\\frac{1}{\\tan(\theta)} = \frac{1+\\tan(3\theta)\\tan(2\theta)}{\\tan(3\theta) - \\tan(2\theta)}\\\\\\\frac{1}{\\tan(\theta)}(\tan(3\theta) - \tan(2\theta)) = 1+\tan(2\theta)\tan(3\theta))