Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}$
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}$
$\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}$
$L.H.S. =\frac{1+\sec A }{\sec A }=\frac{1+\frac{1}{\cos A }}{\frac{1}{\cos A }}$
$=\frac{\frac{\cos A+1}{\cos A}{1}}{\frac{1}{\cos A}}=(\cos A+1)$
$=\frac{(1-\cos A)(1+\cos A)}{(1-\cos A)}$
$=\frac{1-\cos ^{2} A}{1-\cos A}=\frac{\sin ^{2} A}{1-\cos A}$
$= R.H.S.$
Similar Questions
Given $\sec \theta=\frac{13}{12},$ calculate all other trigonometric ratios.
Given $\sec \theta=\frac{13}{12},$ calculate all other trigonometric ratios.
If $\cot \theta=\frac{7}{8},$ evaluate:
$(i)$ $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
$(ii)$ $\cot ^{2} \theta$
If $\cot \theta=\frac{7}{8},$ evaluate:
$(i)$ $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
$(ii)$ $\cot ^{2} \theta$
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$\frac{\tan 26^{\circ}}{\cot 64^{\circ}}$
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$\frac{\tan 26^{\circ}}{\cot 64^{\circ}}$
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Express the trigonometric ratios $\sin A , \sec A$ and $\tan A$ in terms of $\cot A$.