Tan- is not defined when - is equal to

The correct option is D π2 Only tan(2n+1)π2 is not definedHence answer is D ...

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Sin(A+B)Sin(A-B)

tanθ is not d...

Question

$tan θ$ is not defined when $θ$ is equal to

0^{0}

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\frac{π}{4}

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\frac{π}{6}

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\frac{π}{2}

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Solution

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

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

If $θ = {sin}^{- 1} x + {cos}^{- 1} x - {tan}^{- 1} x, x \geq 0$ then the smallest interval in which $θ$ lies is

\frac{π}{4}

θ

\frac{π}{4}

θ

θ

θ \in (\frac{π}{4}, \frac{π}{2})

f (θ) = sec 2 θ - tan 2 θ

f (\frac{π}{4} - θ)

θ

(- \frac{π}{2}, \frac{π}{2})

(1 - t a n θ) (1 + t a n θ) s e c^{2} θ + 2 t a n^{2} θ + 2 = 0

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