Q67 / 180 · KCET 2007 · Mathematics

$\sin ^{2} 17.5^{\circ}+\sin ^{2} 72.5^{\circ}$ is equal to

  • A
    $\cos ^{2} 90^{\circ}$
  • Correct answer
    B
    $\tan ^{2} 45^{\circ}$
  • C
    $\cos ^{2} 30^{\circ}$
  • D
    $\sin ^{2} 45^{\circ}$

Solution

We have, $\sin ^{2} 17.5^{\circ}+\sin ^{2} 72.5^{\circ}$ $$ \begin{aligned} &=\sin ^{2} 17.5^{\circ}+\sin ^{2}\left(90^{\circ}-17.5^{\circ}\right) \\ &=\sin ^{2} 17.5^{\circ}+\cos ^{2} 17.5^{\circ} \\ &=1=1^{2}=\tan ^{2} 45^{\circ} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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