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Trigonometric Ratios of Allied Angles

The absolute ...

Question

The absolute value of $tan \frac{π}{16} + tan \frac{5 π}{16} + tan \frac{9 π}{16} + tan \frac{13 π}{16}$ is

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Solution

Let $θ = \frac{π}{16} \Rightarrow 8 θ = \frac{π}{2}$
$tan \frac{π}{16} + tan \frac{5 π}{16} + tan \frac{9 π}{16} + tan \frac{13 π}{16} = tan θ + tan 5 θ + tan 9 θ + tan 13 θ = tan θ + tan 5 θ + tan (8 θ + θ) + tan (8 θ + 5 θ) = tan θ + tan 5 θ + tan (\frac{π}{2} + θ) + tan (\frac{π}{2} + 5 θ) = tan θ + tan 5 θ - cot θ - cot 5 θ = tan θ - cot θ + tan 5 θ - cot 5 θ (∵ cot θ - tan θ = 2 cot 2 θ) = - 2 cot 2 θ - 2 cot 10 θ = - 2 [cot 2 θ + cot (8 θ + 2 θ)] = - 2 [cot 2 θ + cot (\frac{π}{2} + 2 θ)] = - 2 [cot 2 θ - tan 2 θ] = - 4 cot 4 θ = - 4 cot \frac{π}{4} = - 4$

Therefore, the absolute value of the expression is $4.$

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Trigonometric Ratios of Allied Angles

Standard XII Mathematics

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