The value of the angle tan−1⁡tan⁡65∘−2tan⁡40∘ in degrees is equal to

tan⁡45∘+20∘−2tan⁡40∘1+tan⁡20∘1−tan⁡20∘−4tan⁡20∘1−tan2⁡20∘=1−tan⁡20∘1+tan⁡20∘=tan⁡45∘−20∘=tan⁡25∘∴tan−1⁡tan⁡25∘=25∘

The value of the angle tan−1⁡tan⁡65∘−2tan⁡40∘ in degrees is equal to

tan⁡45∘+20∘−2tan⁡40∘1+tan⁡20∘1−tan⁡20∘−4tan⁡20∘1−tan2⁡20∘=1−tan⁡20∘1+tan⁡20∘=tan⁡45∘−20∘=tan⁡25∘∴tan−1⁡tan⁡25∘=25∘

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The value of the angle $\tan^{- 1} (\tan 65^{\circ} - 2 \tan 40^{\circ})$ in degrees is equal to

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$20^{0}$

$- 20^{0}$

$25^{0}$

$40^{\circ}$

answer is C.

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Detailed Solution

$\begin{aligned} \tan (45^{\circ} + 20^{\circ}) - 2 \tan 40^{\circ} \\ \frac{1 + \tan 20^{\circ}}{1 - \tan 20^{\circ}} - \frac{4 \tan 20^{\circ}}{1 - \tan^{2} 20^{\circ}} \end{aligned}$

$\begin{array}{l} = \frac{(1 - \tan 20^{\circ})}{(1 + \tan 20^{\circ})} \\ = \tan (45^{\circ} - 20^{\circ}) = \tan 25^{\circ} \\ ∴ \tan^{- 1} (\tan 25^{\circ}) = 25^{\circ} \end{array}$