Trigonometrical Ratios and Identities 1 Question 19

19. For 0<θ<π2, the solution(s) of

m=16cosecθ+(m1)π4cosecθ+mπ4=42 is/are

(a) π4

(b) π6

(c) π12

(d) 5π12

(2009)

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Answer:

Correct Answer: 19. (a, b)

Solution:

  1. For 0<θ<π2

m=16cosecθ+(m1)π4cosecθ+mπ4=42m=161sinθ+(m1)π4sinθ+mπ4=42m=16sinθ+mπ4θ+(m1)π4sinπ4sinθ+(m1)π4sinθ+mπ4=42

m=16cotθ+(m1)π4cotθ+mπ41/2=42m=16cotθ+(m1)π4cotθ+mπ4=4cot(θ)cotθ+π4+cotθ+π4cotθ+2π4++cotθ+5π4cotθ+6π4=4cotθcot3π2+θ=4cotθ+tanθ=4tan2θ4tanθ+1=0(tanθ2+3)(tanθ2)23=0tanθ=23 or tanθ=2+3θ=π12;θ=5π12



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