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Question: If \(\tan\alpha = \frac{m}{m + 1}\)and \(\tan\beta = \frac{1}{2m + 1}\), then \(\alpha + \beta =\)...

If tan⁡α=mm+1\tan\alpha = \frac{m}{m + 1}and tan⁡β=12m+1\tan\beta = \frac{1}{2m + 1}, then α+β=\alpha + \beta =

A

π3\frac{\pi}{3}

B

π4\frac{\pi}{4}

C

π6\frac{\pi}{6}

D

None of these

Answer

π4\frac{\pi}{4}

Explanation

Solution

We have, tan⁡α=mm+1\tan\alpha = \frac{m}{m + 1} and tan⁡β=12m+1\tan\beta = \frac{1}{2m + 1}

We know tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}

=mm+1+12m+11−m(m+1)1(2m+1)=2m2+m+m+12m2+m+2m+1−m= \frac{\frac{m}{m + 1} + \frac{1}{2m + 1}}{1 - \frac{m}{(m + 1)}\frac{1}{(2m + 1)}} = \frac{2m^{2} + m + m + 1}{2m^{2} + m + 2m + 1 - m}

=2m2+2m+12m2+2m+1=1⇒tan⁡(α+β)=tan⁡π4= \frac{2m^{2} + 2m + 1}{2m^{2} + 2m + 1} = 1 \Rightarrow \tan(\alpha + \beta) = \tan\frac{\pi}{4}

Hence, α+β=π4\alpha + \beta = \frac{\pi}{4}.

Trick : As α+β\alpha + \beta is independent of m, therefore put m=1,m = 1,

then tan⁡α=12\tan\alpha = \frac{1}{2} and tan⁡β=13\tan\beta = \frac{1}{3}.

Therefore, tan⁡(α+β)=(1/2)+(1/3)1−(1/6)=1.\tan(\alpha + \beta) = \frac{(1/2) + (1/3)}{1 - (1/6)} = 1. Hence α+β=π4.\alpha + \beta = \frac{\pi}{4}.

(Also check for other values of m).