If $\sec \theta = \frac{5}{3}$,then $\tan \theta = \ldots$
- A$1$
- B$\frac{3}{5}$
- C$\frac{4}{3}$
- D$\frac{3}{4}$
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Similar Questions
The value of the expression $\left[\operatorname{cosec}(75^{\circ}+\theta)-\sec(15^{\circ}-\theta)-\tan(55^{\circ}+\theta)+\cot(35^{\circ}-\theta)\right]$ is
Medium
View SolutionIf $1+\sin ^{2} \theta=3 \sin \theta \cos \theta,$ then prove that $\tan \theta=1$ or $\frac{1}{2}$.
Difficult
View SolutionProve that,
$1 + \frac{\cot^{2} \alpha}{1 + \operatorname{cosec} \alpha} = \operatorname{cosec} \alpha$
$1 + \frac{\cot^{2} \alpha}{1 + \operatorname{cosec} \alpha} = \operatorname{cosec} \alpha$
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View Solution$\cot \theta \cdot \tan \theta = \dots$
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View SolutionWrite 'True' or 'False' and justify your answer.
$\frac{\tan 47^{\circ}}{\cot 43^{\circ}}=1$
$\frac{\tan 47^{\circ}}{\cot 43^{\circ}}=1$
Easy
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