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Exercise 4.6 · Q10

Q.tan⁡−1(14)+tan⁡−1(29)\tan^{-1}\left(\dfrac14\right) + \tan^{-1}\left(\dfrac29\right) is equal to

(1) 12cos⁡−1(35)\dfrac12\cos^{-1}\left(\dfrac35\right)
(2) 12sin⁡−1(35)\dfrac12\sin^{-1}\left(\dfrac35\right)
(3) 12tan⁡−1(35)\dfrac12\tan^{-1}\left(\dfrac35\right)
(4) tan⁡−1(12)\tan^{-1}\left(\dfrac12\right)
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The tangent addition formula collapses the sum to tan⁡−1(12)\tan^{-1}\left(\dfrac12\right); converting that through the double-angle identity 2tan⁡−1t=cos⁡−11−t21+t22\tan^{-1}t=\cos^{-1}\dfrac{1-t^2}{1+t^2} (valid for t≥0t\ge0) expresses it as 12cos⁡−1(35)\dfrac12\cos^{-1}\left(\dfrac35\right), matching option (1).

Step 1. Apply the tangent addition formula. With a=14, b=29a=\dfrac14,\ b=\dfrac29: a+b=936+836=1736a+b=\dfrac9{36}+\dfrac8{36}=\dfrac{17}{36}, ab=236=118ab=\dfrac2{36}=\dfrac1{18}, so 1−ab=17181-ab=\dfrac{17}{18}.

Step 2. Compute the ratio. a+b1−ab=17/3617/18=1736⋅1817=12\dfrac{a+b}{1-ab}=\dfrac{17/36}{17/18}=\dfrac{17}{36}\cdot\dfrac{18}{17}=\dfrac12.

Step 3. Conclude the sum. Since ab=118<1ab=\dfrac1{18}<1 and both terms are positive acute angles, tan⁡−114+tan⁡−129=tan⁡−112\tan^{-1}\dfrac14+\tan^{-1}\dfrac29=\tan^{-1}\dfrac12.

Step 4. Convert using the double-angle identity. For t≥0t\ge0, 2tan⁡−1t=cos⁡−11−t21+t22\tan^{-1}t=\cos^{-1}\dfrac{1-t^2}{1+t^2}. With t=12t=\dfrac12: 1−1/41+1/4=3/45/4=35\dfrac{1-1/4}{1+1/4}=\dfrac{3/4}{5/4}=\dfrac35, so 2tan⁡−112=cos⁡−1352\tan^{-1}\dfrac12=\cos^{-1}\dfrac35, i.e. tan⁡−112=12cos⁡−135\tan^{-1}\dfrac12=\dfrac12\cos^{-1}\dfrac35. …

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