If tan35°=k , then the value of tan145°−tan125°1+tan145°tan125°=

tan35=k tan145=tan(180−35) =−tan35 =−k tan125=tan(90+35) =−cot35 =−1k tan145−tan1251+tan145.tan125=−k+1k1+(−k)(−1k)=1−k22(k)

tan35=k tan145=tan(180−35) =−tan35 =−k tan125=tan(90+35) =−cot35 =−1k tan145−tan1251+tan145.tan125=−k+1k1+(−k)(−1k)=1−k22(k)

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

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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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\tan 35 = k

$\tan 145 = \tan (180 - 35)$

$= - \tan 35$

$= - k$

$\tan 125 = \tan (90 + 35)$

$= - \cot 35$

$= - \frac{1}{k}$

$\frac{\tan 145 - \tan 125}{1 + \tan 145. \tan 125} = \frac{- k + \frac{1}{k}}{1 + (- k) (- \frac{1}{k})} = \frac{1 - k^{2}}{2 (k)}$

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