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If tan mθ + cot nθ = 0, then the general value of θ is (a) \(\frac{rπ}{m+n}\)(b) \(\frac{rπ}{m-n}\)(c) \(\frac{(2r+1)π}{(m+n)}\)(d) \(\frac{(2r+1)π}{2(m-n)}\) |
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Answer» Answer : (d) \(\frac{(2r+1)π}{2(m-n)}\) tan mθ + cot nθ = 0 ⇒ tan mθ = – cot nθ ⇒ tan mθ = tan (π/2 + nθ) (∵ tan (π/2 + x) = – cot x) ⇒ mθ = rπ + π/2 + nθ, r∈I ⇒ (m – n)θ = (2r + 1) π/2, r∈I ⇒ θ = \(\frac{(2r+1)π}{2(m-n)}\) , r \(\in\) I |
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