Inverse Trigonometric FunctionshardFree

Inverse Trigonometric Functions: Value Positive Integers Common Factors Find Value

JEE Maths question with a full step-by-step solution.

Question
If sin⁡(30°+arctan⁡x)=1314\sin(30°+\arctan x)=\dfrac{13}{14} and 0<x<10<x<1, the value of xx is a3b\dfrac{a\sqrt{3}}{b}, where a,ba,b are positive integers with no common factors. Find the value of (a+b2)\left(\dfrac{a+b}{2}\right).
Solution
Answer: 8
Step 1: sin⁡(π6+tan⁡−1x)=1314\sin\left(\dfrac{\pi}{6}+\tan^{-1}x\right)=\dfrac{13}{14}. Step 2: Let sin⁡−11314=θ\sin^{-1}\dfrac{13}{14}=\theta. Then sin⁡θ=1314\sin\theta=\dfrac{13}{14}, so by Pythagoras: cos⁡θ=196−16914=2714=3314\cos\theta=\dfrac{\sqrt{196-169}}{14}=\dfrac{\sqrt{27}}{14}=\dfrac{3\sqrt{3}}{14}. Step 3: So tan⁡θ=13/1433/14=1333\tan\theta=\dfrac{13/14}{3\sqrt{3}/14}=\dfrac{13}{3\sqrt{3}}, giving θ=tan⁡−11333\theta=\tan^{-1}\dfrac{13}{3\sqrt{3}}. Step 4: From the equation:
π6+tan⁡−1x=tan⁡−11333  ⇒  tan⁡−1x=tan⁡−11333−tan⁡−113\dfrac{\pi}{6}+\tan^{-1}x=\tan^{-1}\dfrac{13}{3\sqrt{3}}\;\Rightarrow\;\tan^{-1}x=\tan^{-1}\dfrac{13}{3\sqrt{3}}-\tan^{-1}\dfrac{1}{\sqrt{3}}
 (using tan⁡π6=13\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}). Step 5: Apply the difference formula:
tan⁡−1x=tan⁡−11333−131+1333⋅13=tan⁡−113−3331+139=tan⁡−11033229\tan^{-1}x=\tan^{-1}\dfrac{\dfrac{13}{3\sqrt{3}}-\dfrac{1}{\sqrt{3}}}{1+\dfrac{13}{3\sqrt{3}}\cdot\dfrac{1}{\sqrt{3}}}=\tan^{-1}\dfrac{\dfrac{13-3}{3\sqrt{3}}}{1+\dfrac{13}{9}}=\tan^{-1}\dfrac{\dfrac{10}{3\sqrt{3}}}{\dfrac{22}{9}}
 Step 6: Simplify:
=tan⁡−11033⋅922=tan⁡−190663=tan⁡−115113=tan⁡−115333=tan⁡−15311=\tan^{-1}\dfrac{10}{3\sqrt{3}}\cdot\dfrac{9}{22}=\tan^{-1}\dfrac{90}{66\sqrt{3}}=\tan^{-1}\dfrac{15}{11\sqrt{3}}=\tan^{-1}\dfrac{15\sqrt{3}}{33}=\tan^{-1}\dfrac{5\sqrt{3}}{11}
 Step 7: Therefore:
x=5311=a3bx=\dfrac{5\sqrt{3}}{11}=\dfrac{a\sqrt{3}}{b}
 So a=5,b=11a=5,b=11 (coprime). Step 8:
a+b2=5+112=8\dfrac{a+b}{2}=\dfrac{5+11}{2}=8
 Answer: 88
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