Write all the other trigonometric ratios of $\angle A$ in terms of $\sec$ $A$.
Write all the other trigonometric ratios of $\angle A$ in terms of $\sec$ $A$.
We know that,
$\cos A=\frac{1}{\sec A}$
Also, $\sin ^{2} A+\cos ^{2} A=1$
$\sin ^{2} A=1-\cos ^{2} A$
$\sin A=\sqrt{1-\left(\frac{1}{\sec A}\right)^{2}}$
$=\sqrt{\frac{\sec ^{2} A-1}{\sec ^{2} A}}=\frac{\sqrt{\sec ^{2} A-1}}{\sec A}$
$\tan ^{2} A+1=\sec ^{2} A$
$\tan ^{2} A=\sec ^{2} A-1$
$\tan A =\sqrt{\sec ^{2} A -1}$
$\cot A =\frac{\cos A }{\sin A } =\frac{\frac{1}{\sec A}}{\frac{\sqrt{\sec ^{2} A-1}}{\sec A}}$
$=\frac{1}{\sqrt{\sec ^{2} A-1}}$
$\operatorname{cosec} A =\frac{1}{\sin A }=\frac{\sec A }{\sqrt{\sec ^{2} A -1}}$
Similar Questions
If $\tan ( A + B )=\sqrt{3}$ and $\tan ( A - B )=\frac{1}{\sqrt{3}} ; 0^{\circ}< A + B \leq 90^{\circ} ; A > B ,$ find $A$ and $B$
If $\tan ( A + B )=\sqrt{3}$ and $\tan ( A - B )=\frac{1}{\sqrt{3}} ; 0^{\circ}< A + B \leq 90^{\circ} ; A > B ,$ find $A$ and $B$
Show that:
$(i)$ $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}=1$
$(ii)$ $\cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ} \sin 52^{\circ}=0$
Show that:
$(i)$ $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}=1$
$(ii)$ $\cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ} \sin 52^{\circ}=0$
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.
$\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}$
$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=$
$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=$
In $\triangle PQR ,$ right $-$ angled at $Q , PR + QR =25\, cm$ and $PQ =5\, cm .$ Determine the values of $\sin P, \cos P$ and $\tan P$.
In $\triangle PQR ,$ right $-$ angled at $Q , PR + QR =25\, cm$ and $PQ =5\, cm .$ Determine the values of $\sin P, \cos P$ and $\tan P$.