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Introduction of Trigonometry — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsIntroduction of Trigonometry1 Mark
MCQ
$\cos 48^{\circ}-\sin 42^{\circ}=$ ?
  • A
    $1$
  • ✓
    $0$
  • C
    $\frac{1}{2}$
  • D
    None of these
✓

Answer

Correct option: B.
$0$
$\cos 48^{\circ}-\sin 42^{\circ}$
$=\cos \left(90^{\circ}-42^{\circ}\right)-\sin 42^{\circ}$
$=\sin 42^{\circ}-\sin 42^{\circ}=0$
$\left[\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right]$

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