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JEE Advanced 2017 · Trigonometric Functions

Let α\alpha and β\beta be non-zero real numbers such that 2(cos⁡β−cos⁡α)+cos⁡αcos⁡β=12(\cos\beta - \cos\alpha) + \cos\alpha \cos\beta = 1. Then which of the following is/are true?

Answers: A. tan⁡(α2)+3tan⁡(β2)=0\tan\left(\frac{\alpha}{2}\right) + \sqrt{3}\tan\left(\frac{\beta}{2}\right) = 0 · C. tan⁡(α2)−3tan⁡(β2)=0\tan\left(\frac{\alpha}{2}\right) - \sqrt{3}\tan\left(\frac{\beta}{2}\right) = 0

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Written solution

Given: 2(cos⁡β−cos⁡α)+cos⁡αcos⁡β=1\text{Given: } 2(\cos\beta - \cos\alpha) + \cos\alpha \cos\beta = 1

α,β∈R,α,β≠0\alpha, \beta \in \mathbb{R},\quad \alpha, \beta \neq 0

To find: which of the given relations hold\text{To find: which of the given relations hold}

Plan: 1. put t=tan⁡α2, s=tan⁡β2\text{Plan: } 1.\ \text{put } t = \tan\dfrac{\alpha}{2},\ s = \tan\dfrac{\beta}{2}

2. write cos⁡α, cos⁡β in terms of t,s2.\ \text{write } \cos\alpha,\ \cos\beta \text{ in terms of } t, s

3. simplify and factorise3.\ \text{simplify and factorise}

t=tan⁡α2,s=tan⁡β2t = \tan\dfrac{\alpha}{2},\qquad s = \tan\dfrac{\beta}{2}

cos⁡α=1−t21+t2,cos⁡β=1−s21+s2\cos\alpha = \dfrac{1 - t^{2}}{1 + t^{2}},\qquad \cos\beta = \dfrac{1 - s^{2}}{1 + s^{2}}

cos⁡β−cos⁡α=2(t2−s2)(1+t2)(1+s2)\cos\beta - \cos\alpha = \dfrac{2(t^{2} - s^{2})}{(1+t^{2})(1+s^{2})}

2(1−s21+s2−1−t21+t2)+(1−t2)(1−s2)(1+t2)(1+s2)=12\left(\dfrac{1-s^{2}}{1+s^{2}} - \dfrac{1-t^{2}}{1+t^{2}}\right) + \dfrac{(1-t^{2})(1-s^{2})}{(1+t^{2})(1+s^{2})} = 1

4(t2−s2)+(1−t2)(1−s2)=(1+t2)(1+s2)4(t^{2} - s^{2}) + (1-t^{2})(1-s^{2}) = (1+t^{2})(1+s^{2})

=4t2−4s2+1−t2−s2+t2s2= 4t^{2} - 4s^{2} + 1 - t^{2} - s^{2} + t^{2}s^{2}

=1+3t2−5s2+t2s2= 1 + 3t^{2} - 5s^{2} + t^{2}s^{2}

1+3t2−5s2+t2s2=1+t2+s2+t2s21 + 3t^{2} - 5s^{2} + t^{2}s^{2} = 1 + t^{2} + s^{2} + t^{2}s^{2}

3t2−5s2=t2+s23t^{2} - 5s^{2} = t^{2} + s^{2}

2t2=6s2⇒t2=3s22t^{2} = 6s^{2} \Rightarrow t^{2} = 3s^{2}

t2−3s2=0⇒(t−3 s)(t+3 s)=0t^{2} - 3s^{2} = 0 \Rightarrow (t - \sqrt{3}\,s)(t + \sqrt{3}\,s) = 0

t−3 s=0(option 3)ort+3 s=0(option 1)t - \sqrt{3}\,s = 0 \quad \text{(option 3)}\quad \text{or}\quad t + \sqrt{3}\,s = 0 \quad \text{(option 1)}

t=3, s=1⇒cos⁡α=−12, cos⁡β=0t = \sqrt{3},\ s = 1 \Rightarrow \cos\alpha = -\dfrac{1}{2},\ \cos\beta = 0

2(0+12)+0=12\left(0 + \dfrac{1}{2}\right) + 0 = 1

Correct options: (1) and (3)\text{Correct options: (1) and (3)}

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