If tan 350 = k, then the value of tan1450-tan12501+tan1450tan1250 =

tan35∘=k tan145∘-tan125∘1+tan145∘ tan125∘ =tan180°-35°-tan90°+35°1+tan180°-35°·tan90°+35° =-tan35°+cot35°1-tan35° -cot35° =-k+1k1+1 =1-k22k

If tan 350 = k, then the value of tan1450-tan12501+tan1450tan1250 =

tan35∘=k tan145∘-tan125∘1+tan145∘ tan125∘ =tan180°-35°-tan90°+35°1+tan180°-35°·tan90°+35° =-tan35°+cot35°1-tan35° -cot35° =-k+1k1+1 =1-k22k

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If tan $35^{0}$ = k, then the value of $\frac{\tan 145^{0} - \tan 125^{0}}{1 + \tan 145^{0} \tan 125^{0}} =$

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$\frac{2 k}{1 - k^{2}}$

$\frac{2 k}{1 + k^{2}}$

$\frac{1 - k^{2}}{2 k}$

$\frac{1 - k^{2}}{1 + k^{2}}$

answer is C.

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Detailed Solution

$\tan 35^{\circ} = k \frac{\tan 145^{\circ} - \tan 125^{\circ}}{1 + \tan 145^{\circ} \tan 125^{\circ}} = \frac{\tan (180 ° - 35 °) - \tan (90 ° + 35 °)}{1 + \tan (180 ° - 35 °) \cdot \tan (90 ° + 35 °)} = \frac{- \tan 35 ° + \cot 35 °}{1 - \tan 35 ° (- \cot 35 °)} = \frac{- k + \frac{1}{k}}{1 + 1} = \frac{1 - k^{2}}{2 k}$