If $\tan \theta = 1$,then $\sin \theta \cdot \cos \theta = \dots$
- A$2$
- B$1$
- C$\frac{1}{2}$
- D$\frac{1}{4}$
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Prove that,
$(\sqrt{3}+ 1) (3-\cot 30^{\circ})=\tan ^{3} 60^{\circ}-2 \sin 60^{\circ}$
$(\sqrt{3}+ 1) (3-\cot 30^{\circ})=\tan ^{3} 60^{\circ}-2 \sin 60^{\circ}$
Medium
View Solution$(\sin 80^{\circ} + \cos 10^{\circ})(\sin 80^{\circ} - \cos 10^{\circ}) = \ldots \ldots \ldots$
Easy
View SolutionWrite 'True' or 'False' and justify your answer.
$(\tan \theta+2)(2 \tan \theta+1)=5 \tan \theta+\sec ^{2} \theta$
$(\tan \theta+2)(2 \tan \theta+1)=5 \tan \theta+\sec ^{2} \theta$
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View Solution$\tan (90^\circ - \theta) = \ldots \ldots \ldots$
Easy
View SolutionWhich of the following groups truly matches the data of Part $I$ with the data of Part $II$?
Part $I$ Part $II$ $1.$ $\cos(90^\circ - \theta)$ $a.$ $\sec \theta$ $2.$ $\cot(90^\circ - \theta)$ $b.$ $\sin \theta$ $3.$ $\operatorname{cosec}(90^\circ - \theta)$ $c.$ $1$ $d.$ $\tan \theta$
| Part $I$ | Part $II$ |
| $1.$ $\cos(90^\circ - \theta)$ | $a.$ $\sec \theta$ |
| $2.$ $\cot(90^\circ - \theta)$ | $b.$ $\sin \theta$ |
| $3.$ $\operatorname{cosec}(90^\circ - \theta)$ | $c.$ $1$ |
| $d.$ $\tan \theta$ |
Medium
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