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8. Introduction to Trigonometry
easy
$\sec ^{2} \theta+\tan ^{2} \theta=\frac{13}{12},$ હોય તો $\sec ^{4} \theta-\tan ^{4} \theta $ = . . .
A
$\frac{12}{13}$
B
$\frac{13}{12}$
C
$1$
D
$\frac{1}{12}$
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
Similar Questions
જોડકા જોડો.
| $1 .$ $\cos \theta$ | $a.$ $\frac{\cos \theta}{\sin \theta}$ |
| $2.$ $\tan \theta$ | $b.$ $\frac{1}{\operatorname{coses} \theta}$ |
| $3 .$ $\cot \theta$ | $c.$ $\frac{1}{\sec \theta}$ |
| $4.$ $\sin \theta$ | $d.$ $\frac{1}{\cot \theta}$ |
| $e.$ $\sin \theta \cdot \cos \theta$ |
easy
