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JEE Main 201815 AprilEvening ShiftOnline
Exam
JEE Main
Year
2018
Date
15 April
Shift
Evening Shift
Mode
Online
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main 2018 · 15 April · Evening Shift · Online · Mathematics · Trigonometric Equations

Consider the following two statements. Statement p : The value of...

Consider the following two statements. Statement p\boldsymbol{p} : The value of sin⁡120∘\sin 120^{{}^{\circ}} can be divided by taking θ=240∘in\theta=240^{{}^{\circ}} inthe equation 2 \sinθ2=1+sin⁡θ−1−sin⁡θ.\frac{\theta}{2}=\sqrt{1+\sin \theta}-\sqrt{1-\sin \theta} .Statement q\boldsymbol{q} : The angles A,B,CandDA, B, C and D of any quadrilateral ABCDsatisfytheA B C Dsatisfy theequation \cos\left(\frac{1}{2}(A+C)\right)+\cos \left(\frac{1}{2}(B+D)\right)=0 Then the truth values ofp and q$ are respectively.

Options

  1. AF,T\mathrm{F}, \mathrm{T}
  2. BT,T\mathrm{T}, \mathrm{T}
  3. CF, F
  4. DT, F

Correct answer

A. F,T\mathrm{F}, \mathrm{T}

Step-by-step solution

Statement p:sin⁡120∘=cos⁡30∘=32⇒2sin⁡120∘=3p : \sin 120^{{}^{\circ}}=\cos 30^{{}^{\circ}}=\frac{\sqrt{3}}{2} \Rightarrow2 \sin 120^{{}^{\circ}}=\sqrt{3} So, 1+sin⁡240∘−1−sin⁡240∘=1−32−1+32≠3\sqrt{1+\sin 240^{{}^{\circ}}}-\sqrt{1-\sin 240^{{}^{\circ}}}=\sqrt{\frac{1-\sqrt{3}}{2}}-\sqrt{\frac{1+\sqrt{3}}{2}} \neq \sqrt{3} Statement q\mathbf{q} : So, A+B+C+D=2πA+B+C+D=2 \pi⇒A+C2+B+D2=π⇒cos⁡(A+C2)+cos⁡(B+D2)=cos⁡(A+C2)−cos⁡(A+C2)=0\begin{aligned}&\Rightarrow \frac{A+C}{2}+\frac{B+D}{2}=\pi\\ &\Rightarrow \cos \left(\frac{A+C}{2}\right)+\cos \left(\frac{B+D}{2}\right)\\ &=\cos \left(\frac{A+C}{2}\right)-\cos \left(\frac{A+C}{2}\right)=0 \end{aligned}Therefore,statementTherefore, statement pisfalseandstatementis false and statementqistrue.is true.
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