Inverse Trigonometric FunctionshardFree

Inverse Trigonometric Functions — JEE Maths practice question

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

Question

\sin(30°+\arctan x)=\dfrac{13}{14}

and

0<x<1

, the value of

x

\dfrac{a\sqrt{3}}{b}

, where

a,b

are positive integers with no common factors. Find the value of

\left(\dfrac{a+b}{2}\right)

Solution

Answer: 8

Step 1:

\sin\left(\dfrac{\pi}{6}+\tan^{-1}x\right)=\dfrac{13}{14}

. Step 2: Let

\sin^{-1}\dfrac{13}{14}=\theta

. Then

\sin\theta=\dfrac{13}{14}

, so by Pythagoras:

\cos\theta=\dfrac{\sqrt{196-169}}{14}=\dfrac{\sqrt{27}}{14}=\dfrac{3\sqrt{3}}{14}

. Step 3: So

\tan\theta=\dfrac{13/14}{3\sqrt{3}/14}=\dfrac{13}{3\sqrt{3}}

, giving

\theta=\tan^{-1}\dfrac{13}{3\sqrt{3}}

. Step 4: From the equation:

\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\dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}

). Step 5: Apply the difference formula:

\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^{-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=\dfrac{5\sqrt{3}}{11}=\dfrac{a\sqrt{3}}{b}

a=5,b=11

(coprime). Step 8:

\dfrac{a+b}{2}=\dfrac{5+11}{2}=8

Answer:

8

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