1.General solution of tan5θ=cot2θ\tan 5\theta = \cot 2\thetatan5θ=cot2θ isa.θ=nπ7+π14\theta = \frac{n\pi}{7} + \frac{\pi}{14}θ=7nπ+14πb.θ=nπ7+π5\theta = \frac{n\pi}{7} + \frac{\pi}{5}θ=7nπ+5πc.θ=nπ7+π2\theta = \frac{n\pi}{7} + \frac{\pi}{2}θ=7nπ+2πd.θ=nπ7+π3\theta = \frac{n\pi}{7} + \frac{\pi}{3}θ=7nπ+3πLogin to continueOnly logged in users canattempt or see the solution.
