8. Introduction to Trigonometry
easy

$9 \sec ^{2} A-9 \tan ^{2} A=..........$

A

$9$

B

$1$

C

$8$

D

$0$

Solution

$9 \sec ^{2} A-9 \tan ^{2} A$

$=9\left(\sec ^{2} A-\tan ^{2} A\right)$

$=9(1)\left[A s \sec ^{2} A-\tan ^{2} A=1\right]$

$=9$

Standard 10
Mathematics

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