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Trigonometric Ratios of Compound Angles

If tanπ/4+θ+t...

Question

If $tan (\frac{π}{4} + θ) + tan (\frac{π}{4} - θ) = 3$ , then the value of ${tan}^{3} (\frac{π}{4} + θ) + {tan}^{3} (\frac{π}{4} - θ)$ is

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Solution

$tan (\frac{π}{4} + θ) + tan (\frac{π}{4} - θ) = 3$
Cubing both the sides
${[tan (\frac{π}{4} + θ) + tan (\frac{π}{4} - θ)]}^{3} = 27 \Rightarrow {tan}^{3} (\frac{π}{4} + θ) + {tan}^{3} (\frac{π}{4} - θ) + 3 tan (\frac{π}{4} + θ) \cdot tan (\frac{π}{4} - θ) [tan (\frac{π}{4} + θ) + tan (\frac{π}{4} - θ)] = 27 \Rightarrow {tan}^{3} (\frac{π}{4} + θ) + {tan}^{3} (\frac{π}{4} - θ) + 3 \cdot 3 = 27 [∵ tan (\frac{π}{4} + θ) \cdot tan (\frac{π}{4} - θ) = 1] ∴ {tan}^{3} (\frac{π}{4} + θ) + {tan}^{3} (\frac{π}{4} - θ) = 18$

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Trigonometric Ratios of Compound Angles

Standard XII Mathematics

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