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8. Introduction to Trigonometry
easy
$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}=........$
A
$\sec ^{2} A$
B
$-1$
C
$\cot ^{2} A$
D
$\tan ^{2} A$
Solution
$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}=\frac{1+\frac{\sin ^{2} A}{\cos ^{2} A}}{1+\frac{\cos ^{2} A}{\sin ^{2} A}}$
$=\frac{\frac{\cos ^{2} A+\sin ^{2} A}{\cos ^{2} A}}{\frac{\sin ^{2} A+\cos ^{2} A}{\sin ^{2} A}}=\frac{\frac{1}{\cos ^{2} A}}{\frac{1}{\sin ^{2} A}}$
$=\frac{\sin ^{2} A }{\cos ^{2} A }=\tan ^{2} A$
Standard 10
Mathematics
