If tan-4-tan-4-3- then the value of tan3-4-tan3-4- is

Tan(π4+θ)+tan(π4 θ)=3 Cubing both the sides [tan(π4+θ)+tan(π4 θ)]^3=27 ⇒ nbsp;tan^3(π4+θ)+tan^3(π4 θ) +3tan(π4+θ)·tan(π4 θ)[tan(π4+θ)+tan(π4 θ)]=27 ⇒tan^3(π4+ ...

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Standard XIII

Mathematics

Trigonometric Ratios of Compound Angles

If tanπ4+θ+ta...

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 XIII Mathematics

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If tan(π4+θ)+tan(π4−θ)=3, then the value of tan3(π4+θ)+tan3(π4−θ) is

tan(π4+θ)+tan(π4−θ)=3 Cubing both the sides [tan(π4+θ)+tan(π4−θ)]3=27⇒tan3(π4+θ)+tan3(π4−θ)+3tan(π4+θ)⋅tan(π4−θ)[tan(π4+θ)+tan(π4−θ)]=27⇒tan3(π4+θ)+tan3(π4−θ)+3⋅3=27[∵tan(π4+θ)⋅tan(π4−θ)=1]∴tan3(π4+θ)+tan3(π4−θ)=18

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