Example 14 - Chapter 3 Class 11 Trigonometric Functions
Last updated at April 16, 2024 by Teachoo
Examples
Example 2 Important
Example 3
Example 4
Example 5 Important
Example 6 Important
Example 7 Important
Example 8
Example 9 Important
Example 10
Example 11 Important
Example 12
Example 13
Example 14 You are here
Example 15
Example 16 Important
Example 17 Important
Example 18 Important
Example 19
Example 20 Important
Example 21 Important
Example 22 Important
Question 1
Question 2
Question 3
Question 4
Question 5 Important
Question 6
Question 7 Important
Last updated at April 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