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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

If x= 15^ ^ ^ , y=cosec 75^ ^ ^ and z=4 18^ ^ ^ , then :

If x=tan⁡15∘,y=cosec⁡75∘x=\tan 15^{{}^{{}^{\circ}}}, y=\operatorname{cosec} 75^{{}^{{}^{\circ}}} and z=4sin⁡18∘z=4 \sin 18^{{}^{{}^{\circ}}}, then :

Options

  1. Ax<y<zx < y < z
  2. By<z<xy < z < x
  3. Cz<x<yz < x < y
  4. Dx<z<yx < z < y

Correct answer

A. x<y<zx < y < z

Step-by-step solution

∵x=tan⁡15∘=tan⁡(45∘−30∘)=tan⁡45∘−tan⁡30∘1+tan⁡45∘tan⁡30∘=1−131+13=3−13+1=(3−1)23−1=3+1−232=2−3\because \quad x=\tan 15^{{}^{{}^{\circ}}}=\tan\left(45^{{}^{{}^{\circ}}}-30^{{}^{{}^{\circ}}}\right)=\frac{\tan 45^{{}^{{}^{\circ}}}-\tan 30^{{}^{{}^{\circ}}}}{1+\tan 45^{{}^{{}^{\circ}}} \tan 30^{{}^{{}^{\circ}}}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)^2}{3-1}=\frac{3+1-2 \sqrt{3}}{2}=2-\sqrt{3} and y=cosec⁡75∘=1sin⁡(45∘+30∘)=1sin⁡45∘cos⁡30∘+cos⁡45∘sin⁡30∘=1322+122=223+1×3−13−1=22(3−1)3−1=6−2y=\operatorname{cosec} 75^{{}^{{}^{\circ}}}=\frac{1}{\sin \left(45^{{}^{{}^{\circ}}}+30^{{}^{{}^{\circ}}}\right)}=\frac{1}{\sin 45^{{}^{{}^{\circ}}} \cos 30^{{}^{{}^{\circ}}}+\cos 45^{{}^{{}^{\circ}}} \sin 30^{{}^{{}^{\circ}}}}=\frac{1}{\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}}=\frac{2 \sqrt{2}}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{2 \sqrt{2}(\sqrt{3}-1)}{3-1}=\sqrt{6}-\sqrt{2} and z=4sin⁡18∘=4\quad z=4 \sin 18^{{}^{{}^{\circ}}}=4\left(\frac{\sqrt{5}-1}{4}\right)=\sqrt{5}-1ItisclearfromabovethatIt is clear from above that(2-\sqrt{3}) < (\sqrt{6}-\sqrt{2}) < (\sqrt{5}-1)\Rightarrow \quad x < y < z
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