Questions
If , then is equal to
a
2kk-1,k∈R
b
2kk-1,k∈[1/3,3]
c
2kk-1,k∉[1/3,3]
d
k-12k,k∉[1/3,3]
detailed solution
Correct option is C
We have, tan3AtanA=k⇒3-tan2A1-3tan2A=k⇒tan2A=k-33k-1Now, sin3AsinA=3−4sin2A=3−41+cot2A=3−41+3k−1k−3=2kk−1Again, sin3AsinA=3−4sin2A⇒ 2kk−1=3−4sin2A⇒ 4sin2A=3−2kk−1⇒ sin2A=k−34(k−1)⇒ 0≤k−34(k−1)≤1 ∵0≤sin2A≤1⇒ k<13 or k>3Hence, sin3AsinA=2kk−1, where k<13 or, k>3.Talk to our academic expert!
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If then is equal to
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