CIE/GCE/AS/Math/P1/20/Mar/12/Q#5
Question
Solve the equation
tan 𝜃 + 3 sin 𝜃 + 2
=2
tan 𝜃 − 3 sin 𝜃 + 1
for 0𝑜 ≤ 𝜃 ≤ 90𝑜.
Solution
We are given the equation;
tan 𝜃 + 3 sin 𝜃 + 2
=2
tan 𝜃 − 3 sin 𝜃 + 1
tan 𝜃 + 3 sin 𝜃 + 2 = 2 tan 𝜃 − 6 sin 𝜃 + 2
9 sin 𝜃 − tan 𝜃 = 0
9 sin 𝜃 − tan 𝜃 = 0
sin 𝜃
Since tan θ = , therefore;
cos 𝜃
sin 𝜃
9 sin 𝜃 − =0
cos 𝜃
9 sin 𝜃 cos 𝜃 − sin 𝜃 = 0
sin 𝜃 (9 cos 𝜃 − 1) = 0
Now we have two options.
Question
Solve the equation
tan 𝜃 + 3 sin 𝜃 + 2
=2
tan 𝜃 − 3 sin 𝜃 + 1
for 0𝑜 ≤ 𝜃 ≤ 90𝑜.
Solution
We are given the equation;
tan 𝜃 + 3 sin 𝜃 + 2
=2
tan 𝜃 − 3 sin 𝜃 + 1
tan 𝜃 + 3 sin 𝜃 + 2 = 2 tan 𝜃 − 6 sin 𝜃 + 2
9 sin 𝜃 − tan 𝜃 = 0
9 sin 𝜃 − tan 𝜃 = 0
sin 𝜃
Since tan θ = , therefore;
cos 𝜃
sin 𝜃
9 sin 𝜃 − =0
cos 𝜃
9 sin 𝜃 cos 𝜃 − sin 𝜃 = 0
sin 𝜃 (9 cos 𝜃 − 1) = 0
Now we have two options.
