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Q.Prove that: 2tan⁡−1(13)+cos⁡−1(35)=π22\tan^{-1}\left(\dfrac13\right)+\cos^{-1}\left(\dfrac35\right)=\dfrac{\pi}{2}.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2025Subjective· 3mImportance★★★★★
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Compute tan⁡\tan of twice the first angle using the double-angle formula, then show it equals cot⁡\cot of the second angle.

Let A=tan⁡−113A=\tan^{-1}\dfrac13, so tan⁡A=13\tan A=\dfrac13, with A∈(0,π4)A\in\left(0,\dfrac\pi4\right) (acute, small).

tan⁡(2A)=2tan⁡A1−tan⁡2A=2⋅131−19=2389=23⋅98=34\tan(2A)=\frac{2\tan A}{1-\tan^2A}=\frac{2\cdot\frac13}{1-\frac19}=\frac{\frac23}{\frac89}=\frac23\cdot\frac98=\frac34

Since 0<A<π/40<A<\pi/4, we have 0<2A<π/20<2A<\pi/2, so 2A=tan⁡−1342A=\tan^{-1}\dfrac34.

Now let B=cos⁡−135B=\cos^{-1}\dfrac35, so cos⁡B=35\cos B=\dfrac35, B∈(0,π/2)B\in(0,\pi/2), giving sin⁡B=1−925=45\sin B=\sqrt{1-\frac9{25}}=\dfrac45 and tan⁡B=43\tan B=\dfrac43.

We want to show 2A+B=π22A+B=\dfrac\pi2, i.e. 2A=π2−B2A=\dfrac\pi2-B. Since both 2A2A and π2−B\dfrac\pi2-B lie in (0,π2)\left(0,\dfrac\pi2\right), it suffices to check their tangents agree: …

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