Sec ^{2} θ+tan ^{2} θ=frac{13}{12}, then value of sec ^{4} θ-tan ^{4} θ is .........

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8. Introduction to Trigonometry

easy

$\sec ^{2} \theta+\tan ^{2} \theta=\frac{13}{12},$ then the value of $\sec ^{4} \theta-\tan ^{4} \theta $ is .........

$\frac{12}{13}$

$\frac{13}{12}$

$1$

$\frac{1}{12}$

Solution

$\sec ^{4} \theta-\tan ^{4} \theta=\left(\sec ^{2} \theta-\tan ^{2} \theta\right)\left(\sec ^{2} \theta+\tan ^{2} \theta\right)=(1)\left(\frac{13}{12}\right)=\frac{13}{12}$

Standard 10

Mathematics

If $\tan ^{2} \theta=\sin ^{2} \theta+\cos ^{2} \theta,$ then $\theta=\ldots \ldots \ldots$

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$\sin ^{2} 60-\tan 45+\cos ^{2} 30-\cot 90=\ldots \ldots \ldots \ldots$

easy

$\sec ^{2} \theta+\tan ^{2} \theta=\frac{13}{12},$ then the value of $\sec ^{4} \theta-\tan ^{4} \theta $ is .........

Solution

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Given that $\alpha+\beta=90^{\circ},$ show that

$\sqrt{\cos \alpha \operatorname{cosec} \beta-\cos \alpha \sin \beta}=\sin \alpha$