Sin θ=frac{3}{5}, then tan θ=ldots ldots ldots ldots

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8. Introduction to Trigonometry

easy

$\sin \theta=\frac{3}{5},$ then $\tan \theta=\ldots \ldots \ldots \ldots$

$\frac{4}{3}$

$\frac{5}{4}$

$\frac{4}{5}$

$\frac{3}{4}$

Solution

$\cos ^{2} \theta=1-\sin ^{2} \theta=1-\left(\frac{3}{5}\right)^{2}=1-\frac{9}{25}=\frac{16}{25}=\left(\frac{4}{5}\right)^{2} \quad \therefore \cos \theta=\frac{4}{5}$

Now, $\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{3 / 5}{4 / 5}=\frac{3}{4}$

Standard 10

Mathematics

$\sin ^{2} 35+\cos ^{2} \theta=1,$ then $\theta=\ldots \ldots \ldots \ldots$

easy

$5 \cos A=4 \sin A,$ then $\tan A=\ldots \ldots \cdots \cdots$

easy

If $\sin \theta+\cos \theta=p$ and $\sec \theta+\operatorname{cosec} \theta=q,$ then prove that $q\left(p^{2}-1\right)=2 p$

hard

$\sin ^{2} 1+\sin ^{2} 3+\sin ^{2} 87+\sin ^{2} 89=\ldots \ldots \ldots \ldots$

easy

$0 < \theta < 90$ and $\sec \theta=\operatorname{cosec} 60,$ then the value of $2 \cos ^{2} \theta-1$ is ……..

medium

$\sin \theta=\frac{3}{5},$ then $\tan \theta=\ldots \ldots \ldots \ldots$

Solution

Similar Questions

$\sin ^{2} 35+\cos ^{2} \theta=1,$ then $\theta=\ldots \ldots \ldots \ldots$

$5 \cos A=4 \sin A,$ then $\tan A=\ldots \ldots \cdots \cdots$

If $\sin \theta+\cos \theta=p$ and $\sec \theta+\operatorname{cosec} \theta=q,$ then prove that $q\left(p^{2}-1\right)=2 p$

$\sin ^{2} 1+\sin ^{2} 3+\sin ^{2} 87+\sin ^{2} 89=\ldots \ldots \ldots \ldots$

$0 < \theta < 90$ and $\sec \theta=\operatorname{cosec} 60,$ then the value of $2 \cos ^{2} \theta-1$ is ……..

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