If $\tan A = \cot B$,prove that $A + B = 90^{\circ}$.
(N/A) Given that,$\tan A = \cot B$.
We know that $\cot B = \tan(90^{\circ} - B)$.
Substituting this into the given equation,we get $\tan A = \tan(90^{\circ} - B)$.
Since the tangent functions are equal,their angles must be equal: $A = 90^{\circ} - B$.
Rearranging the terms,we get $A + B = 90^{\circ}$.
We know that $\cot B = \tan(90^{\circ} - B)$.
Substituting this into the given equation,we get $\tan A = \tan(90^{\circ} - B)$.
Since the tangent functions are equal,their angles must be equal: $A = 90^{\circ} - B$.
Rearranging the terms,we get $A + B = 90^{\circ}$.
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$(i)$ $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} = 1$
$(ii)$ $\cos 38^{\circ} \cos 52^{\circ} - \sin 38^{\circ} \sin 52^{\circ} = 0$
$(i)$ $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} = 1$
$(ii)$ $\cos 38^{\circ} \cos 52^{\circ} - \sin 38^{\circ} \sin 52^{\circ} = 0$
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$\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}$
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View SolutionGiven $15 \cot A = 8$,find $\sin A$ and $\sec A$.
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Medium
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