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Standard X

Mathematics

Relation between Trigonometric Ratios

2 tan 30∘ 1-t...

Question

$\frac{2 \tan 30 °}{1 - \tan^{2} 30 °}$ is equal to

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

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Solution

We are asked to find the value of the following

We know that

Hence the correct option is

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Similar questions

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 °

If cos 2x = cos 60^o cos 30^o + sin 60^o sin 30^o then x is equal to

Q. Question 2 (iv)
Choose the correct option and justify your choice.
(iv)

\frac{2 t a n 3 0^{0}}{1 - t a n^{2} 30^{0}} =

(A)

c o s 60^{\circ}

(B)

s i n 60^{\circ}

(C)

t a n 60^{\circ}

(D)

s i n 30^{\circ}

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2 tan 30°1-tan2 30° is equal to (a) cos 60° (b) sin 60° (c) tan 60° (d) sin 30°

We are asked to find the value of the following We know that Hence the correct option is

$\frac{2 \tan 30 °}{1 - \tan^{2} 30 °}$ is equal to

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

We are asked to find the value of the following

We know that

Hence the correct option is