Prove that \tan [2 \tan ^{-1} \frac{1}{2}-\cot ^{-1} 3] =\frac{9}{13}

CBSE Class 12 Math 2020 Delhi Set 2 Solved Paper - Question 9

Question : 9 of 11

Marks: +1, -0

Prove that

\tan [2 \tan ^{-1} \;{1}/{2}-\cot ^{-1} 3] =\;{9}/{13}

Solution:

Let

m=\tan ^{-1} \;{1}/{2}

⇒ \;\; \tan m=\;{1}/{2} \;\;\;⋅⋅⋅⋅⋅⋅⋅(i)

and

n=\cot ^{-1} 3

⇒ \;\; \cot n=3 .

⇒ \;\; \tan n=\;{1}/{3} \;\;\;⋅⋅⋅⋅⋅⋅⋅(ii)

∴ \tan [2 \tan ^{-1} \;{1}/{2}-\cot ^{-1} 3] =\tan (2 m-n)

=\;{\tan 2 m-\tan n}/{1+\tan 2 m \tan n .}

{[\table ∵ \tan (a-b) ; =\;{\tan a-\tan b}/{1+\tan a \tan b}]}

=\;{\;{2 \tan m}/{1-\tan ^2 m}-\tan n}/{1+\;{2 \tan m}/{1-\tan ^2 m} \tan n}

=\;{\;{2 × 1 / 2}/{1- ({1}/{2}) ^2}-{1}/{3}}/{1+\;{2 × 1 / 2}/{1- (\;{1}/{2}) ^2} × \;{1}/{3}}

=\;{1}/{1-\;{1}/{4}-\;{1}/{3}}\;=\;{\;{4}/{3}-{1}/{3}}/{1+\;{1}/{1-\;{1}/{4}} × \;{1}/{3}} × \;{1}/{3}

\;=\;{1}/{1+\;{4}/{9}}

\;=\;{9}/{9+4}

\;=\;{9}/{13} .

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