Example 14 - Chapter 3 Class 11 Trigonometric Functions
Last updated at Dec. 16, 2024 by Teachoo
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Last updated at Dec. 16, 2024 by Teachoo
Example 14 Show that tan 3π₯ tan 2π₯ tan π₯ = tan 3π₯ β tan 2π₯ β tan π₯ We know that ππ = ππ+ π Therefore, tan ππ = πππβ‘(ππ + π) tanβ‘3π₯ = tanβ‘γ2π₯ +γ tanγβ‘π₯ γ/(1βtanβ‘γ2π₯ tanβ‘π₯ γ ) " " tanβ‘3π₯βtanβ‘3π₯ tanβ‘2π₯ tanβ‘π₯=tanβ‘2π₯+tanβ‘π₯ tanβ‘3π₯βtanβ‘2π₯βtanβ‘π₯=tanβ‘3π₯ tanβ‘2π₯ tanβ‘π₯ πππβ‘ππ πππβ‘ππ πππβ‘π = πππβ‘ππ β πππβ‘ππ β πππβ‘π