Tan θ+cot θ=ldots ldots ldots

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8. Introduction to Trigonometry

medium

$\tan \theta+\cot \theta=\ldots \ldots \ldots$

$2$

$\sin \theta$

$\cos \theta$

$\operatorname{cosec} \theta \cdot \sec \theta$

Solution

$\tan \theta+\cot \theta=\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}=\frac{1}{\sin \theta \cos \theta}=\operatorname{cosec} \theta \cdot \sec \theta$

Standard 10

Mathematics

$\frac{\cos ^{2} 40+\cos ^{2} 50}{\sin ^{2} 40+\sin ^{2} 50}=\ldots \ldots \ldots \ldots$

easy

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

easy

લઘૂકોણ $\theta$ માટે જો $\cos \theta=\sin \theta,$ તો $2 \tan ^{2} \theta+\sin ^{2} \theta+1=\ldots \ldots \ldots \ldots$

medium

$\operatorname{cosec} A=\sqrt{10},$ તો $\sin A=\ldots \ldots \ldots . . .$

easy

$\sin 70=\cos \theta,$ તો $\theta=\ldots \ldots \ldots \ldots$

easy

$\tan \theta+\cot \theta=\ldots \ldots \ldots$

Solution

Similar Questions

$\frac{\cos ^{2} 40+\cos ^{2} 50}{\sin ^{2} 40+\sin ^{2} 50}=\ldots \ldots \ldots \ldots$

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

લઘૂકોણ $\theta$ માટે જો $\cos \theta=\sin \theta,$ તો $2 \tan ^{2} \theta+\sin ^{2} \theta+1=\ldots \ldots \ldots \ldots$

$\operatorname{cosec} A=\sqrt{10},$ તો $\sin A=\ldots \ldots \ldots . . .$

$\sin 70=\cos \theta,$ તો $\theta=\ldots \ldots \ldots \ldots$

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