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Sign of Trigonometric Ratios in Different Quadrants

If tan3 π/11+...

Question

If $t a n (\frac{3 π}{11}) + 4 s i n (\frac{2 π}{11}) = λ$ then the value of $λ^{2} - 9$ is equal to ___

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Solution

$λ = \frac{1}{c o s (\frac{3 π}{14})} {s i n (\frac{3 π}{11}) + 4 s i n (\frac{2 π}{11}) c o s (\frac{3 π}{11})} λ c o s (\frac{3 π}{11}) = s i n (\frac{3 π}{11}) + 4 s i n (\frac{2 π}{11}) c o s (\frac{3 π}{11}) λ^{2} c o s^{2} (\frac{3 π}{11}) = s i n^{2} (\frac{3 π}{11}) + 16 s i n^{2} (\frac{2 π}{11}) c o s^{2} (\frac{3 π}{11}) + 8 s i n (\frac{2 π}{11}) s i n (\frac{3 π}{11}) c o s (\frac{3 π}{11}) λ^{2} (2 c o s^{2} (\frac{3 π}{11})) = 2 s i n^{2} (\frac{3 π}{11}) + 32 s i n^{2} (\frac{2 π}{11}) c o s^{2} (\frac{3 π}{11}) + 8 s i n (\frac{2 π}{11}) s i n (\frac{6 π}{11})$

$λ^{2} (2 c o s^{2} (\frac{3 π}{11})) = (1 - c o s (\frac{6 π}{11})) + 8 (1 - c o s \frac{4 π}{11}) (1 + c o s \frac{6 π}{11}) + 4 (c o s \frac{4 π}{11} - c o s \frac{8 π}{11})$

$λ^{2} (2 c o s^{2} (\frac{3 π}{11})) = 9 + 7 c o s (\frac{6 π}{11}) - 4 c o s (\frac{4 π}{11}) - 8 c o s (\frac{4 π}{11}) c o s (\frac{6 π}{11}) - 4 c o s (\frac{8 π}{11})$

$λ^{2} (2 c o s^{2} (\frac{3 π}{11})) = 9 + 7 c o s (\frac{6 π}{11}) - 4 c o s (\frac{4 π}{11}) - 4 (c o s (\frac{10 π}{11}) + c o s (\frac{2 π}{11})) - 4 c o s (\frac{8 π}{11})$

$λ^{2} (2 c o s^{2} (\frac{3 π}{11})) = 9 + 11 c o s (\frac{6 π}{11}) - 4 {c o s (\frac{2 π}{11}) + c o s (\frac{4 π}{11}) + c o s (\frac{6 π}{11}) + c o s (\frac{8 π}{11}) + c o s (\frac{10 π}{11})}$

$λ^{2} (2 c o s^{2} (\frac{3 π}{11})) = 9 + 11 c o s (\frac{6 π}{11}) - 4 (\frac{s i n π - s i n \frac{π}{11}}{2 s i n \frac{π}{11}})$

$λ^{2} (2 c o s^{2} (\frac{3 π}{11})) = 9 + 11 c o s (\frac{6 π}{11}) + 2$

$λ^{2} (2 c o s^{2} (\frac{3 π}{11})) = 11 (1 + c o s (\frac{6 π}{11}))$
$λ^{2} (2 c o s^{2} (\frac{3 π}{11})) = 11 (2 c o s^{2} (\frac{3}{11}))$

$λ^{2} = 11$
$∴ λ^{2} - 9 = 11 - 9 = 2$

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