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If $\mathrm{A}+\mathrm{B}=90^{\circ}$, then what is $\sqrt{\sin \mathrm{A} \sec \mathrm{B}-\sin \mathrm{A} \cos \mathrm{B}}$
$\begin{array}{ll}\text { equal to ? }
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$\begin{array}{ll}\text { equal to ? }
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The correct answer is:
$\cos A$
Let $\mathrm{A}+\mathrm{B}=90^{\circ}$
Consider $\sqrt{\sin A \sec B-\sin A \cos B}$
$\begin{aligned} &=\sqrt{\sin \mathrm{A} \sec \left(90^{\circ}-\mathrm{A}\right)-\sin \mathrm{A} \cos \left(90^{\circ}-\mathrm{A}\right)} \\ &=\sqrt{\sin \mathrm{A} \operatorname{cosec} \mathrm{A}-\sin \mathrm{A} \sin \mathrm{A}}=\sqrt{1-\sin ^{2} \mathrm{~A}}=\cos \mathrm{A} . \\ \text { 140. (a) } & \text { Consider } \tan ^{4} \mathrm{~A}-\sec ^{4} \mathrm{~A}+\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A} \\ &=\left(\tan ^{2} \mathrm{~A}\right)^{2}-\left(\sec ^{2} \mathrm{~A}\right)^{2}+\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A} . \\ &=\left(\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A}\right)\left(\tan ^{2} \mathrm{~A}-\sec ^{2} \mathrm{~A}\right) \\ &=\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A} \\ &=-\left(\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A}\right)+\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A}=0 . \end{aligned}$
Consider $\sqrt{\sin A \sec B-\sin A \cos B}$
$\begin{aligned} &=\sqrt{\sin \mathrm{A} \sec \left(90^{\circ}-\mathrm{A}\right)-\sin \mathrm{A} \cos \left(90^{\circ}-\mathrm{A}\right)} \\ &=\sqrt{\sin \mathrm{A} \operatorname{cosec} \mathrm{A}-\sin \mathrm{A} \sin \mathrm{A}}=\sqrt{1-\sin ^{2} \mathrm{~A}}=\cos \mathrm{A} . \\ \text { 140. (a) } & \text { Consider } \tan ^{4} \mathrm{~A}-\sec ^{4} \mathrm{~A}+\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A} \\ &=\left(\tan ^{2} \mathrm{~A}\right)^{2}-\left(\sec ^{2} \mathrm{~A}\right)^{2}+\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A} . \\ &=\left(\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A}\right)\left(\tan ^{2} \mathrm{~A}-\sec ^{2} \mathrm{~A}\right) \\ &=\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A} \\ &=-\left(\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A}\right)+\tan ^{2} \mathrm{~A}+\sec ^{2} \mathrm{~A}=0 . \end{aligned}$
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