The principal argument of tan-i is

The correct option is B θ cis(θ π/2) z=tanθ i =1cosθ(sinθ icosθ) =secθ(cos(θ π2)+isin(θ π2)) Hence |z|=secθ and arg(z)=θ π2 ...

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Sum of Trigonometric Ratios in Terms of Their Product

The principal...

Question

The principal argument of $(tan θ - i)$ is

cos θ c i s (\frac{π}{2} - θ)

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sec θ c i s (θ - \frac{π}{2})

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cos θ c i s (\frac{π}{2} + θ)

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sec θ c i s (\begin{matrix} {π - -}_{+ θ} 2 \end{matrix})

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Similar questions

tan θ - i,

θ \in (- \frac{π}{2}, \frac{π}{2})

r e^{i β}

(r, β)

tan θ = \frac{1}{\sqrt{15}}

0 < θ < \frac{π}{2}

\frac{c s c^{2} θ - {sec}^{2} θ}{c s c^{2} θ + {sec}^{2} θ}

cos θ = \frac{3}{5}, π < θ < \frac{3 π}{2}

\frac{csc θ + cot θ}{sec θ - tan θ}

θ, 0 < θ < \frac{π}{2},

6 \sum m = 1 c o s e c [θ + \frac{(m - 1) π}{4}] c o s e c [θ + \frac{m π}{4}] = 4 \sqrt{2}

\frac{π}{12}

\frac{5 π}{12}

tan θ + cot θ = 4, 0 < θ < \frac{π}{2},

θ = \frac{π}{12}

\frac{5 π}{12}

l i m θ \to \frac{π}{2} \frac{s e c θ - t a n θ}{π - 2 θ} =

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