1 Trigonometric Equations
If 4nα = π, then the value of tanα. tan2α. tan3α. tan4α. ….. tan(2n – 2)α tan (2n – 1)α is 1) 0 2) 1 3)-1 4) 1/2
Solution
2nα = π/2
(tanα tan(2n – 1)α) (tan 2α tan(2n – 2)α) ….. (tan (n – 1)α tan (n + 1)α).tan nα
(tanα tan(π/2 - α))(tan2αtan(π/2-2α))......tanπ/4 =1
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