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CBSE Class 12 Math 2020 Delhi Set 2 Solved Paper

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Question : 9 of 11
Marks: +1, -0
Prove that tan⁡[2tan⁡−112−cot⁡−13]=913\tan \left[ 2 \tan^{-1} \frac{1}{2} - \cot^{-1} 3 \right] = \frac{9}{13}
Solution:  
Let m=tan⁡−112m = \tan^{-1} \frac{1}{2} .
⇒tan⁡m=12⋯(i)\Rightarrow \quad \tan m = \frac{1}{2} \quad \cdots (i)
and n=cot⁡−13n = \cot^{-1} 3
⇒cot⁡n=3.\Rightarrow \quad \cot n = 3 .
⇒tan⁡n=13⋯(ii)\Rightarrow \quad \tan n = \frac{1}{3} \quad \cdots (ii)
∴tan⁡[2tan⁡−112−cot⁡−13]=tan⁡(2m−n)\therefore \tan \left[ 2 \tan^{-1} \frac{1}{2} - \cot^{-1} 3 \right] = \tan (2m - n)
=tan⁡2m−tan⁡n1+tan⁡2mtan⁡n.= \frac{\tan 2m - \tan n}{1 + \tan 2m \tan n}. [∵tan⁡(a−b)=tan⁡a−tan⁡b1+tan⁡atan⁡b]\left[ \begin{array}{l} \because \tan(a-b) \\ = \frac{\tan a - \tan b}{1 + \tan a \tan b} \end{array} \right]
=2tan⁡m1−tan⁡2m−tan⁡n1+2tan⁡m1−tan⁡2mtan⁡n= \frac{ \frac{2 \tan m}{1 - \tan^2 m} - \tan n }{ 1 + \frac{2 \tan m}{1 - \tan^2 m} \tan n }
 
=2×121−(12)2−131+2×121−(12)2×13= \frac{ \frac{2 \times \frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2} - \frac{1}{3} }{ 1 + \frac{2 \times \frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2} \times \frac{1}{3} }
=11−14−13=43−131+11−14×13×13= \frac{1}{1 - \frac{1}{4} - \frac{1}{3}} = \frac{ \frac{4}{3} - \frac{1}{3} }{ 1 + \frac{1}{1 - \frac{1}{4}} \times \frac{1}{3} } \times \frac{1}{3}
=11+49= \frac{1}{1 + \frac{4}{9}}
=99+4= \frac{9}{9+4}
=913.= \frac{9}{13} .
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