The value of 2 tan 30 - 1 - tan 2 30 - is

The correct option is B tan 60 ^ ∘ ⇒ (2tan30^01 tan^230^0 )=tan(2× 30^0) =tan60^0 ........ >formula>used (2tanθ1 tan^2θ )=tan2θ ...

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Standard XII

Mathematics

Circular Measurement of Angle

The value of ...

Question

The value of $\frac{2 tan 30^{\circ}}{1 - {tan}^{2} 30^{\circ}}$ is

cos 60^{\circ}

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sin 60^{\circ}

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tan 60^{\circ}

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sin 30^{\circ}

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Solution

The correct option is B $tan 60^{\circ}$
$\Rightarrow$ $(\frac{2 t a n 30^{0}}{1 - t a n^{2} 30^{0}}) = tan (2 \times 30^{0})$

$= t a n 60^{0}$ ........ $f o r m u l a u s e d (\frac{2 t a n θ}{1 - t a n^{2} θ}) = t a n 2 θ$

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Similar questions

\frac{2 \tan 30 °}{1 - \tan^{2} 30 °}

is equal to

(a) cos 60°
(b) sin 60°
(c) tan 60°
(d) sin 30°

Choose the correct option and justify your choice.

(i)

(A). sin60°

(B). cos60°

(C). tan60°

(D). sin30°

(ii)

(A). tan90°

(B). 1

(C). sin45°

(D). 0

(iii) sin2A = 2sinA is true when A =

(A). 0°

(B). 30°

(C). 45°

(D). 60°

(iv)

(A). cos60°

(B). sin60°

(C). tan60°

(D). sin30°

Show that:

(i) $\frac{1 - sin 60^{\circ}}{cos 60^{\circ}} = \frac{tan 60^{\circ} - 1}{tan 60^{\circ} + 1}$

(ii) $\frac{cos 30^{\circ} + sin 60^{\circ}}{1 + sin 30^{\circ} + cos 60^{\circ}} = cos 30^{\circ}$

Q. Show that:
(i)

\frac{1 - \sin 60 °}{\cos 60 °} = \frac{\tan 60 ° - 1}{\tan 60 ° + 1}

(ii)

\frac{\cos 30 ° + \sin 60 °}{1 + \sin 30 ° + \cos 60 °} = \cos 30 °

Q. If

x tan 30^{\circ} = \frac{sin 30^{\circ} + cos 60^{\circ}}{tan 60^{\circ} + sin 60^{\circ}}

, then the value of

x

is:

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The value of 2tan30∘1−tan230∘ is

The correct option is B tan60∘⇒(2tan3001−tan2300)=tan(2×300)=tan600 ........formulaused(2tanθ1−tan2θ)=tan2θ

The value of $\frac{2 tan 30^{\circ}}{1 - {tan}^{2} 30^{\circ}}$ is

The correct option is B $tan 60^{\circ}$
$\Rightarrow$ $(\frac{2 t a n 30^{0}}{1 - t a n^{2} 30^{0}}) = tan (2 \times 30^{0})$

$= t a n 60^{0}$ ........ $f o r m u l a u s e d (\frac{2 t a n θ}{1 - t a n^{2} θ}) = t a n 2 θ$