$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=$
$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=$
- A
$\tan 90^{\circ}$
- B
$1$
- C
$0$
- D
$\sin 45^{\circ}$
Similar Questions
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Evaluate:
$\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}$
Evaluate:
$\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}$
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In $\triangle$ $PQR,$ right-angled at $Q$ (see $Fig.$), $PQ =3 \,cm$ and $PR =6 \,cm$. Determine $\angle QPR$ and $\angle PRQ$.
In $\triangle$ $PQR,$ right-angled at $Q$ (see $Fig.$), $PQ =3 \,cm$ and $PR =6 \,cm$. Determine $\angle QPR$ and $\angle PRQ$.