NDA MATHEMATICS QUIZ - 23

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Question 1:

$cosec θ - 2 \cot 2 θ \cdot \cos θ = . . . . . . .$

2 \cos θ

2 \cos θ

\sin θ

\sin θ

2 \sin θ

2 \sin θ

2 \sin θ \cdot \cos θ

2 \sin θ \cdot \cos θ

$\frac{4 \cos 2 x}{(\sin 2 x)^{2}}$ is equal to.

$\frac{4 \cos 2 x}{(\sin 2 x)^{2}}$ = . . . . . .

\cot^{2} x + \tan^{2} x

\cot^{2} x + \tan^{2} x

\cot^{2} x - \tan^{2} x

\cot^{2} x - \tan^{2} x

2 \cot^{2} x - 3 \tan^{2} x

2 \cot^{2} x - 3 \tan^{2} x

None of these

Which of the following is correct?

\tan 70^{\circ} = \tan 50^{\circ} + 2 \tan 20^{\circ}

\tan 70^{\circ} = \tan 50^{\circ} + 2 \tan 20^{\circ}

\tan 70^{\circ} = 2 \tan 50^{\circ} + \tan 20^{\circ}

\tan 70^{\circ} = 2 \tan 50^{\circ} + \tan 20^{\circ}

\tan 70^{\circ} = 2 \tan 50^{\circ} + 3 \tan 20^{\circ}

\tan 70^{\circ} = 2 \tan 50^{\circ} + 3 \tan 20^{\circ}

\tan 70^{\circ} = 3 \tan 50^{\circ} + 2 \tan 20^{\circ}

\tan 70^{\circ} = 3 \tan 50^{\circ} + 2 \tan 20^{\circ}

Value of $\cos^{4} \frac{π}{8} + \cos^{4} \frac{3 π}{8} + \cos^{4} \frac{5 π}{8} +$ $\cos^{4} \frac{7 π}{8}$ is equal to.

$\cos^{4} \frac{π}{8} + \cos^{4} \frac{3 π}{8} + \cos^{4} \frac{5 π}{8} +$ $\cos^{4} \frac{7 π}{8}$ किसके बराबर है ?

\frac{1}{2}

\frac{1}{2}

\frac{3}{2}

\frac{3}{2}

\frac{1}{8}

\frac{1}{8}

\frac{3}{16}

\frac{3}{16}

$\cos^{2} \frac{π}{9} + \cos^{2} \frac{2 π}{9} + \cos^{2} \frac{4 π}{9}$ is equal to.

$\cos^{2} \frac{π}{9} + \cos^{2} \frac{2 π}{9} + \cos^{2} \frac{4 π}{9}$ किसके बराबर है ?

\frac{1}{2}

\frac{1}{2}

\frac{5}{2}

\frac{5}{2}

\frac{7}{2}

\frac{7}{2}

\frac{3}{2}

\frac{3}{2}

$(1 + \cos \frac{π}{8}) (1 + \cos \frac{3 π}{8}) (1 + \cos \frac{5 π}{8})$ $(1 + \cos \frac{7 π}{8}) = . . . . . . . .$

\frac{1}{6}

\frac{1}{6}

\frac{1}{8}

\frac{1}{8}

\frac{1}{16}

\frac{1}{16}

\frac{1}{2}

\frac{1}{2}

If $\cot θ = (\frac{b}{a})$ then $(a \cos 2 θ + b \sin 2 θ)$ is independent of

अगर $\cot θ = (\frac{b}{a})$ b हो तो $(a \cos 2 θ + b \sin 2 θ)$ निम्न में से किससे मुक्त है

b

b

a

a

Both a & b

Neither b nor a

If $\sin 25^{\circ} + \cos 25^{\circ} = λ$ then $\cos 20^{\circ}$ will be equal to

अगर $\sin 25^{\circ} + \cos 25^{\circ} = λ$ तो $\cos 20^{\circ}$ किसके बराबर होगा ?

2 λ

2 λ

\sqrt{2} λ

\sqrt{2} λ

\frac{λ}{\sqrt{2}}

\frac{λ}{\sqrt{2}}

2 λ^{2}

2 λ^{2}

$\cos \frac{6 π}{7} + \cos \frac{4 π}{7} + \cos \frac{2 π}{7} = . . . . . .$

\frac{1}{2}

\frac{1}{2}

\frac{1}{8}

\frac{1}{8}

\frac{1}{4}

\frac{1}{4}

- \frac{1}{2}

- \frac{1}{2}

Value of $\frac{1 - 4 \sin 10^{\circ} \cdot \sin 70^{\circ}}{2 \sin 10^{\circ}}$ equals.

$\frac{1 - 4 \sin 10^{\circ} \cdot \sin 70^{\circ}}{2 \sin 10^{\circ}}$ का मान किसके बराबर होगा ?