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8. Introduction to Trigonometry
medium
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}$
Option A
Option B
Option C
Option D
Solution
$\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}$
$L.H.S. =\frac{1+\sec A }{\sec A }=\frac{1+\frac{1}{\cos A }}{\frac{1}{\cos A }}$
$=\frac{\frac{\cos A+1}{\cos A}{1}}{\frac{1}{\cos A}}=(\cos A+1)$
$=\frac{(1-\cos A)(1+\cos A)}{(1-\cos A)}$
$=\frac{1-\cos ^{2} A}{1-\cos A}=\frac{\sin ^{2} A}{1-\cos A}$
$= R.H.S.$
Standard 10
Mathematics
