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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2016 · Mathematics · Trigonometric Ratios & Identities

If A= ( 12^ - 36^ ) ( 96^ + 24^ ) and B=( 60^ - 12^ )( 48^ - 72^ )...

If A=A=\left(\cos 12^{\circ}-\cos 36^{\circ}\right)(sin⁡96∘+sin⁡24∘)\left(\sin 96^{\circ}+\sin 24^{\circ}\right) and B=(sin⁡60∘−sin⁡12∘)(cos⁡48∘−cos⁡72∘)\mathrm{B}=\left(\sin 60^{\circ}-\sin 12^{\circ}\right)\left(\cos 48^{\circ}-\cos 72^{\circ}\right), then what is AB\frac{\mathrm{A}}{\mathrm{B}} equal to?

Options

  1. A–1
  2. B0
  3. C1
  4. D2

Correct answer

C. 1

Step-by-step solution

 Given A=(cos⁡12∘−cos⁡36∘)(sin⁡96∘+sin⁡24∘)B=(sin⁡60∘−sin⁡12∘)(cos⁡48∘−cos⁡72∘)AB=[2sin⁡24∘sin⁡12∘][2sin⁡60∘cos⁡36∘][2cos⁡36∘sin⁡24∘][2sin⁡60∘sin⁡12∘]⇒AB=1\begin{aligned} & \text{ Given } A=\left(\cos 12^{\circ}-\cos 36^{\circ}\right)\left(\sin 96^{\circ}+\sin 24^{\circ}\right) \\ & B=\left(\sin 60^{\circ}-\sin 12^{\circ}\right)\left(\cos 48^{\circ}-\cos 72^{\circ}\right) \\ & \frac{A}{B}=\frac{\left[2 \sin 24^{\circ} \sin 12^{\circ}\right]\left[2 \sin 60^{\circ} \cos 36^{\circ}\right]}{\left[2 \cos 36^{\circ} \sin 24^{\circ}\right]\left[2 \sin 60^{\circ} \sin 12^{\circ}\right]} \\ & \Rightarrow \frac{A}{B}=1\end{aligned}
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