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Question: Find the value of \(\tan \left( {\dfrac{\pi }{4} + \dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right)...

Find the value of tan⁡(π4+12cos⁡−1ab)+tan⁡(π4−12cos⁡−1ab)\tan \left( {\dfrac{\pi }{4} + \dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right) + \tan \left( {\dfrac{\pi }{4} - \dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right)
(A). 2ab (B). ab (C). ba (D). 2ba  \left( A \right).{\text{ }}\dfrac{{2a}}{b} \\\ \left( B \right).{\text{ }}\dfrac{a}{b} \\\ \left( C \right).{\text{ }}\dfrac{b}{a} \\\ \left( D \right).{\text{ }}\dfrac{{2b}}{a} \\\

Explanation

Solution

Hint: Solve by using simple trigonometric identities of tan⁡θ\tan \theta and cos⁡θ\cos \theta .

Given tan⁡(π4+12cos⁡−1ab)+tan⁡(π4−12cos⁡−1ab) …(1)\tan \left( {\dfrac{\pi }{4} + \dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right) + \tan \left( {\dfrac{\pi }{4} - \dfrac{1}{2}{{\cos }^{ - 1}}\dfrac{a}{b}} \right){\text{ }} \ldots \left( 1 \right)
Let 12cos⁡−1ab=θ\dfrac{1}{2}{\cos ^{ - 1}}\dfrac{a}{b} = \theta
∴cos⁡−1ab=2θ cos⁡2θ=ab …(2)  \therefore {\cos ^{ - 1}}\dfrac{a}{b} = 2\theta \\\ \cos 2\theta = \dfrac{a}{b}{\text{ }} \ldots \left( 2 \right) \\\
Put the value of 12cos⁡−1ab=θ\dfrac{1}{2}{\cos ^{ - 1}}\dfrac{a}{b} = \theta in equation (1)\left( 1 \right), we get
⇒tan⁡(π4+θ)+tan⁡(π4−θ) …(3)\Rightarrow \tan \left( {\dfrac{\pi }{4} + \theta } \right) + \tan \left( {\dfrac{\pi }{4} - \theta } \right){\text{ }} \ldots \left( 3 \right)
We know that, tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}and tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}
Using above identities in(3)\left( 3 \right), we get
⇒tan⁡π4+tan⁡θ1−tan⁡π4tan⁡θ+tan⁡π4−tan⁡θ1+tan⁡π4tan⁡θ\Rightarrow \dfrac{{\tan \dfrac{\pi }{4} + \tan \theta }}{{1 - \tan \dfrac{\pi }{4}\tan \theta }} + \dfrac{{\tan \dfrac{\pi }{4} - \tan \theta }}{{1 + \tan \dfrac{\pi }{4}\tan \theta }}
Putting the value of tan⁡π4=1\tan \dfrac{\pi }{4} = 1 in above equation, we get
⇒1+tan⁡θ1−1tan⁡θ+1−tan⁡θ1+1tan⁡θ ⇒1+tan⁡θ1−tan⁡θ+1−tan⁡θ1+tan⁡θ  \Rightarrow \dfrac{{1 + \tan \theta }}{{1 - 1\tan \theta }} + \dfrac{{1 - \tan \theta }}{{1 + 1\tan \theta }} \\\ \Rightarrow \dfrac{{1 + \tan \theta }}{{1 - \tan \theta }} + \dfrac{{1 - \tan \theta }}{{1 + \tan \theta }} \\\
Using cross multiplication, we get

⇒(1+tan⁡θ)2+(1−tan⁡θ)2(1−tan⁡θ)(1+tan⁡θ) ⇒1+2tan⁡θ+tan⁡2θ+1−2tan⁡θ+tan⁡2θ(1)2−(tan⁡θ)2 [∵(a+b)2=a2+2ab+b2 and (a−b)2=a2−2ab+b2] ⇒2+2tan⁡2θ1−tan⁡2θ  \Rightarrow \dfrac{{{{\left( {1 + \tan \theta } \right)}^2} + {{\left( {1 - \tan \theta } \right)}^2}}}{{\left( {1 - \tan \theta } \right)\left( {1 + \tan \theta } \right)}} \\\ \Rightarrow \dfrac{{1 + 2\tan \theta + {{\tan }^2}\theta + 1 - 2\tan \theta + {{\tan }^2}\theta }}{{{{\left( 1 \right)}^2} - {{\left( {\tan \theta } \right)}^2}}}{\text{ }}\left[ {\because {{\left( {a + b} \right)}^2} = {a^2} + 2ab + {b^2}{\text{ and }}{{\left( {a - b} \right)}^2} = {a^2} - 2ab + {b^2}} \right] \\\ \Rightarrow \dfrac{{2 + 2{{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }} \\\

Taking 22 common from numerator, we get
⇒2(1+tan⁡2θ1−tan⁡2θ)\Rightarrow 2\left( {\dfrac{{1 + {{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }}} \right)
We know that cos⁡2θ=1−tan⁡2θ1+tan⁡2θ\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}. Hence, we can write above equation as:
⇒2(cos⁡2θ)\Rightarrow \dfrac{2}{{\left( {\cos 2\theta } \right)}}
Now, putting the value of cos⁡2θ\cos 2\theta from equation(2)\left( 2 \right)in above equation, we get
⇒2(ab) ⇒2ba  \Rightarrow \dfrac{2}{{\left( {\dfrac{a}{b}} \right)}} \\\ \Rightarrow \dfrac{{2b}}{a} \\\
∴\therefore Correct option is (D)\left( D \right).

Note: In these types of problems, one should always try to convert the equation to some trigonometric identity by either taking out the common terms from the equations or try to minimize it by using the identities.