Trigonometrical Ratios and Identities 2 Question 1
1. If $\alpha+\beta=\frac{\pi}{2}$ and $\beta+\gamma=\alpha$, then $\tan \alpha$ equals $\quad(2001,1 M)$
(a) $2(\tan \beta+\tan \gamma)$
(b) $\tan \beta+\tan \gamma$
(c) $\tan \beta+2 \tan \gamma$
(d) $2 \tan \beta+\tan \gamma$
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Answer:
Correct Answer: 1. (c)
Solution:
- Given,
$$ \begin{array}{lc} \Rightarrow & \alpha=(\pi / 2)-\beta \\ \Rightarrow & \tan \alpha=\tan (\pi / 2-\beta) \\ \Rightarrow & \tan \alpha=\cot \beta \\ \Rightarrow & \tan \alpha \tan \beta=1 \\ \text { Again, } & \beta+\gamma=\alpha \\ \Rightarrow & \gamma=(\alpha-\beta) \\ \Rightarrow & \tan \gamma=\tan (\alpha-\beta) \\ \Rightarrow & \tan \gamma=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta} \\ \Rightarrow & \tan \gamma=\frac{\tan \alpha-\tan \beta}{1+1} \\ \therefore & 2 \tan \gamma=\tan \alpha-\tan \beta \\ \Rightarrow & \tan \alpha=\tan \beta+2 \tan \gamma \end{array} $$
