Question

Answer: (4)

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Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

6   (that's option 4)

Step-by-step explanation:

The functions here have periods of length π, so it's enough to work out the answer just for an interval of size π; the answer for the interval (0, 3π) will then just be 3 times that.  [ Note that this means we can already rule out "4" being the answer! ]

Consider the interval (π/4, 5π/4).  Over this interval, the LHS ranges from -∞ to +∞, always increasing from one asymptote to the next.  [ It's just tan(x) shifted left by π/4. ]

Meanwhile, the RHS starts at  cot(π/4) - 2 = 1-2 = -1 and returns to this same value, always decreasing and going via an asymptote at π.

Now:

• from π/4 to π, LHS goes from -∞ to 1 always increasing, while RHS goes from -1 to -∞ always decreasing, so the two functions cross at a unique  point in this interval;
• from π to 5π/4, LHS goes from 1 to +∞ always increasing, while RHS goes from +∞ to -1 always decreasing, so again the two functions cross at a unique point in this interval.

So the two sides are equal exactly twice in an interval of size π.

Therefore the two sides are equal exactly 6 times in the interval (0, 3π).