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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

If the value of the expression tan ((1)/(2) ^ - 1 2 5 ) is in the...

If the value of the expression tantan\left(\dfrac{1}{2}\cos^{- 1} \dfrac{2}{\sqrt{5}}\right)isintheformofis in the form ofa + \sqrt{b}wherewherea, b \in Z, then the value of a+bb\dfrac{a + b}{b} is K, Find 10K.

Correct answer

6

Step-by-step solution

We know that,

tan⁡(x2)=1−cosxsinx\tan \left(\dfrac{x}{2}\right) = \dfrac{1 - cos x}{sin x}

∴tan⁡(12cos⁡−125)=1−cos⁡(cos⁡−1(25))sin⁡(cos⁡−1(25))∴ \tan \left(\dfrac{1}{2}\cos^{- 1} \dfrac{2}{\sqrt{5}}\right) = \dfrac{1 - \cos \left(\cos^{- 1} \left(\dfrac{2}{\sqrt{5}}\right)\right)}{\sin \left(\cos^{- 1} \left(\dfrac{2}{\sqrt{5}}\right)\right)}

=1−2/5sin⁡(sin⁡−1(1/5))= \dfrac{1 - 2 / \sqrt{5}}{\sin \left(\sin^{- 1} \left(1 / \sqrt{5}\right)\right)}

=1−2/51/5=5−2= \dfrac{1 - 2 / \sqrt{5}}{1 / \sqrt{5}} = \sqrt{5} - 2

Now, a+b=−2+5a + \sqrt{b} = - 2 + \sqrt{5}

⇒a=−2⇒ a = - 2 and b=5b = 5

Hence, a+bb=35=0.6\dfrac{a + b}{b} = \dfrac{3}{5} = 0.6
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