Sec θ=frac{5}{3}, તો tan θ=ldots ldots ldots ldots

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8. Introduction to Trigonometry

easy

$\sec \theta=\frac{5}{3},$ તો $\tan \theta=\ldots \ldots \ldots \ldots$

$1$

$\frac{3}{5}$

$\frac{4}{3}$

$\frac{3}{4}$

Solution

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

$\therefore \tan \theta=\frac{4}{3}$

Standard 10

Mathematics

$\tan 23\,\,\tan 42\,\, \tan 48\,\, \tan 67=\ldots \ldots \ldots \ldots .$

hard

$\left(\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}\right)$ નું મૂલ્ય ……….. છે.

medium

$\sec (90-\theta)=\ldots \ldots \ldots$

easy

$\left[\operatorname{cosec}\left(75^{\circ}+\theta\right)-\sec \left(15^{\circ}-\theta\right)-\tan \left(55^{\circ}+\theta\right)+\right.\left.\cot \left(35^{\circ}-\theta\right)\right]$ નું મૂલ્ય ……….. છે.

medium

$2 \sin ^{2} \theta+4 \sec ^{2} \theta+5 \cot ^{2} \theta+2 \cos ^{2} \theta-4 \tan ^{2} \theta-5 \operatorname{cosec}^{2} \theta=\ldots \ldots \ldots$

medium

$\sec \theta=\frac{5}{3},$ તો $\tan \theta=\ldots \ldots \ldots \ldots$

Solution

Similar Questions

$\tan 23\,\,\tan 42\,\, \tan 48\,\, \tan 67=\ldots \ldots \ldots \ldots .$

$\left(\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}\right)$ નું મૂલ્ય ……….. છે.

$\sec (90-\theta)=\ldots \ldots \ldots$

$\left[\operatorname{cosec}\left(75^{\circ}+\theta\right)-\sec \left(15^{\circ}-\theta\right)-\tan \left(55^{\circ}+\theta\right)+\right.\left.\cot \left(35^{\circ}-\theta\right)\right]$ નું મૂલ્ય ……….. છે.

$2 \sin ^{2} \theta+4 \sec ^{2} \theta+5 \cot ^{2} \theta+2 \cos ^{2} \theta-4 \tan ^{2} \theta-5 \operatorname{cosec}^{2} \theta=\ldots \ldots \ldots$

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