Frac{sec θ-1}{sec θ+1}=ldots ldots ldots ldots

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8. Introduction to Trigonometry

easy

$\frac{\sec \theta-1}{\sec \theta+1}=\ldots \ldots \ldots \ldots$

$\tan \theta$

$\frac{1-\sin \theta}{1+\sin \theta}$

$\frac{\operatorname{cosec} \theta-1}{\operatorname{cosec} \theta+1}$

$\frac{1-\cos \theta}{1+\cos \theta}$

Solution

$\frac{\sec \theta-1}{\sec \theta+1}=\frac{(\sec \theta-1) \times \cos \theta}{(\sec \theta+1) \times \cos \theta}=\frac{1-\cos \theta}{1+\cos \theta}$

Standard 10

Mathematics

$\sin ^{2} 15+\sin ^{2} 75=\ldots \ldots \ldots \ldots$

easy

Prove that,

$\tan \theta+\tan \left(90^{\circ}-\theta\right)=\sec \theta \sec \left(90^{\circ}-\theta\right)$

medium

$\cot \theta=\frac{a}{b},$ then $\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=\ldots \ldots \ldots$

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

Prove that:

If $\tan A =\frac{3}{4},$ then $\sin A \cos A =\frac{12}{25}$

medium

$\frac{\sec \theta-1}{\sec \theta+1}=\ldots \ldots \ldots \ldots$

Solution

Similar Questions

$\sin ^{2} 15+\sin ^{2} 75=\ldots \ldots \ldots \ldots$

Prove that, $\tan \theta+\tan \left(90^{\circ}-\theta\right)=\sec \theta \sec \left(90^{\circ}-\theta\right)$

$\cot \theta=\frac{a}{b},$ then $\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=\ldots \ldots \ldots$

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

Prove that: If $\tan A =\frac{3}{4},$ then $\sin A \cos A =\frac{12}{25}$

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Prove that,

$\tan \theta+\tan \left(90^{\circ}-\theta\right)=\sec \theta \sec \left(90^{\circ}-\theta\right)$

Prove that:

If $\tan A =\frac{3}{4},$ then $\sin A \cos A =\frac{12}{25}$