If tan 45- - - then the value of - in radians is

The correct option is C π4 tan 45^∘ = θ tanθ = θ #160; #160;only for θ = π/ 4 . Hence, θ = 45^∘ or π/ 4 . ...

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Applications of Cross Product

If tan 45∘ ...

Question

If $tan 45^{\circ} = cot θ$ , then the value of $θ$ , in radians is

π

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\frac{π}{9}

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\frac{π}{4}

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\frac{π}{12}

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^{'} a^{'}

^{'} b^{'}

f (x)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x + a \sqrt{2} sin x, & 0 \leq x < \frac{π}{4} 2 x cot x + b, & \frac{π}{4} \leq x \leq \frac{π}{2} a cos 2 x - b sin x, & \frac{π}{2} < x \leq π \end{matrix}

x = \frac{π}{4}, \frac{π}{2}

\frac{π}{4}

θ

\frac{π}{4}

θ

θ

tan (45^{\circ} + θ) = \sqrt{3}

0 < θ < 20^{\circ}

θ

tan [\frac{π}{4} + θ] + tan [\frac{π}{4} - θ] = α

{tan}^{3} [\frac{π}{4} + θ] + {tan}^{3} [\frac{π}{4} - 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}

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