If tan A = cot(2A - 30°), then what is the value of A?A30°B25°C18°D40°Answer: D. 40° Read Explanation: Given:tanA=cot(2A−30∘)\tan A = \cot(2A - 30^\circ)tanA=cot(2A−30∘)Use the identity:cotθ=tan(90∘−θ)\cot \theta = \tan(90^\circ - \theta)cotθ=tan(90∘−θ)So,tanA=tan(90∘−(2A−30∘))\tan A = \tan\big(90^\circ - (2A - 30^\circ)\big)tanA=tan(90∘−(2A−30∘))tanA=tan(120∘−2A)\tan A = \tan(120^\circ - 2A)tanA=tan(120∘−2A)For (tanx=tany),(\tan x = \tan y),(tanx=tany),x=y+n⋅180∘x = y + n \cdot 180^\circx=y+n⋅180∘Thus,A=120∘−2A+n⋅180∘A = 120^\circ - 2A + n \cdot 180^\circA=120∘−2A+n⋅180∘3A=120∘+n⋅180∘3A = 120^\circ + n \cdot 180^\circ3A=120∘+n⋅180∘A=40∘+n⋅60∘A = 40^\circ + n \cdot 60^\circA=40∘+n⋅60∘The principal acute-angle solution is:40∘\boxed{40^\circ}40∘ Read more in App