$(\sin 80^{\circ} + \cos 10^{\circ})(\sin 80^{\circ} - \cos 10^{\circ}) = \ldots \ldots \ldots$
- A$1$
- B$2$
- C$0$
- D$\cos^{2} 10^{\circ}$
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Similar Questions
In right-angled $\Delta ABC$,$m\angle B = 90^\circ$. If $AC = 20$ and $m\angle C = 30^\circ$,then $BC = \ldots$
Medium
View SolutionEvaluate the following:
$2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}$
$2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}$
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View Solution$0 < \theta < 90$ and $\sec \theta = \operatorname{cosec} 60^\circ$,then the value of $2 \cos^2 \theta - 1$ is ........
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View SolutionIf $\tan \theta = \sqrt{3}$,then $\theta = \ldots$ (in $^\circ$)
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View SolutionWrite 'True' or 'False' and justify your answer.
$\frac{\tan 47^{\circ}}{\cot 43^{\circ}}=1$
$\frac{\tan 47^{\circ}}{\cot 43^{\circ}}=1$
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