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Trigonometric Functions and Equations question

2017 · Shift 2 · Q29
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  5. /2017 · Shift 2 · Q29

Trigonometric Functions and Equations question

2017 · Shift 2 · Q29

JEE AdvancedMathematicsTrigonometric Functions and EquationsMultiple correct+4 / −2
Let α\alphaα and β\betaβ be non zero real numbers such that 2(cos⁡β−cos⁡α)+cos⁡αcos⁡β=12(\cos \beta - \cos \alpha ) + \cos \alpha \cos \beta = 12(cosβ−cosα)+cosαcosβ=1. Then which of the following is/are true?
  1. A
    3tan⁡(α2)−tan⁡(β2)=2\sqrt 3 \tan \left( {{\alpha \over 2}} \right) - \tan \left( {{\beta \over 2}} \right) = 23​tan(2α​)−tan(2β​)=2
  2. B
    tan⁡(α2)−3tan⁡(β2)=0\tan \left( {{\alpha \over 2}} \right) - \sqrt 3 \tan \left( {{\beta \over 2}} \right) = 0tan(2α​)−3​tan(2β​)=0
  3. C
    tan⁡(α2)+3tan⁡(β2)=0\tan \left( {{\alpha \over 2}} \right) + \sqrt 3 \tan \left( {{\beta \over 2}} \right) = 0tan(2α​)+3​tan(2β​)=0
  4. D
    3tan⁡(α2)+tan⁡(β2)=2\sqrt 3 \tan \left( {{\alpha \over 2}} \right) + \tan \left( {{\beta \over 2}} \right) = 23​tan(2α​)+tan(2β​)=2
View written solutionFree

Correct answer: B, C

  1. We are given 2(cos⁡β−cos⁡α)+cos⁡αcos⁡β=1.2(\cos\beta-\cos\alpha)+\cos\alpha\cos\beta=1.2(cosβ−cosα)+cosαcosβ=1. We want to convert this into an equation involving tan⁡α2,tan⁡β2.\tan\frac\alpha2,\quad \tan\frac\beta2.tan2α​,tan2β​.

  2. Use the tangent half-angle substitution: cos⁡α=1−t21+t2,cos⁡β=1−u21+u2,\cos\alpha=\frac{1-t^2}{1+t^2},\qquad \cos\beta=\frac{1-u^2}{1+u^2},cosα=1+t21−t2​,cosβ=1+u21−u2​, where t=tan⁡α2,u=tan⁡β2.t=\tan\frac\alpha2,\qquad u=\tan\frac\beta2.t=tan2α​,u=tan2β​.

Then the given equation becomes 2(1−u21+u2−1−t21+t2)+(1−t2)(1−u2)(1+t2)(1+u2)=1.2\left(\frac{1-u^2}{1+u^2}-\frac{1-t^2}{1+t^2}\right)+\frac{(1-t^2)(1-u^2)}{(1+t^2)(1+u^2)}=1.2(1+u21−u2​−1+t21−t2​)+(1+t2)(1+u2)(1−t2)(1−u2)​=1.

  1. Take LCM (1+t2)(1+u2)(1+t^2)(1+u^2)(1+t2)(1+u2).

First,

=\frac{(1-u^2)(1+t^2)-(1-t^2)(1+u^2)}{(1+t^2)(1+u^2)}.$$ Expand the numerator: $$(1-u^2)(1+t^2)=1+t^2-u^2-t^2u^2,$$ $$(1-t^2)(1+u^2)=1+u^2-t^2-t^2u^2.$$ So the difference is $$1+t^2-u^2-t^2u^2-(1+u^2-t^2-t^2u^2)=2t^2-2u^2=2(t^2-u^2).$$ Hence $$2\left(\frac{1-u^2}{1+u^2}-\frac{1-t^2}{1+t^2}\right)=\frac{4(t^2-u^2)}{(1+t^2)(1+u^2)}.$$ Also, $$\cos\alpha\cos\beta=\frac{(1-t^2)(1-u^2)}{(1+t^2)(1+u^2)}.$$ Thus the equation becomes $$\frac{4(t^2-u^2)+(1-t^2)(1-u^2)}{(1+t^2)(1+u^2)}=1.$$ Multiply through: $$4(t^2-u^2)+(1-t^2)(1-u^2)=(1+t^2)(1+u^2).$$ 4. Expand both sides. Left side: $$4t^2-4u^2+1-t^2-u^2+t^2u^2=1+3t^2-5u^2+t^2u^2.$$ Right side: $$1+t^2+u^2+t^2u^2.$$ Equating: $$1+3t^2-5u^2+t^2u^2=1+t^2+u^2+t^2u^2.$$ Cancel common terms $1$ and $t^2u^2$: $$3t^2-5u^2=t^2+u^2.$$ So, $$2t^2-6u^2=0$$ $$t^2=3u^2.$$ Hence $$t=\pm \sqrt3\,u.$$ That is, $$\tan\frac\alpha2=\pm \sqrt3\tan\frac\beta2.$$ 5. Now check the options. Since $$t=\sqrt3 u \quad \text{or} \quad t=-\sqrt3 u,$$ we get two possible relations: - If $t=\sqrt3 u$, then $$t-\sqrt3 u=0,$$ which is option **B**. - If $t=-\sqrt3 u$, then $$t+\sqrt3 u=0,$$ which is option **C**. 6. Check A and D. Option A is $$\sqrt3 t-u=2.$$ Using $t=\pm\sqrt3 u$: - if $t=\sqrt3 u$, then $\sqrt3 t-u=3u-u=2u$, not always $2$; - if $t=-\sqrt3 u$, then $\sqrt3 t-u=-3u-u=-4u$, not always $2$. So **A is not necessarily true**. Option D is $$\sqrt3 t+u=2.$$ Using $t=\pm\sqrt3 u$: - if $t=\sqrt3 u$, then $\sqrt3 t+u=3u+u=4u$, not always $2$; - if $t=-\sqrt3 u$, then $\sqrt3 t+u=-3u+u=-2u$, not always $2$. So **D is not necessarily true**. 7. Therefore the correct options are $$\boxed{B,\ C}.$$
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