Consider the following statements : 1. If $ an{\alpha = \frac{m}{m + 1}, an{\beta = \frac{1}{2m + 1},\ then\ \alpha + \beta = \frac{\pi}{4}}}$ 2. If $3 an{( heta - 15{^\circ}) = an{( heta + 15{^\circ}),\ 0

Question

Consider the following statements : 1. If $	an{\alpha = \frac{m}{m + 1},	an{\beta = \frac{1}{2m + 1},\ then\ \alpha + \beta = \frac{\pi}{4}}}$ 2. If $3	an{(	heta - 15{^\circ}) = 	an{(	heta + 15{^\circ}),\ 0 < 	heta < \pi,\ then\ 	heta = \frac{\pi}{4}}}$ 3. If $\sin^{2}{ax - \sin^{2}{(a - 1)x = \sin^{2}{x,}}}$then $x$ is equal to $\frac{n\pi}{a - 1}$ Which of the statements given above are correct?

Dr. Abubacker · Accepted Answer

(a)
(1) $	an{(\alpha + \beta) = \frac{	an{\alpha + 	an\beta}}{1 - 	an{\alpha	an\beta}}}$
$= \frac{\frac{m}{m + 1} + \frac{1}{2m + 1}}{1 - \frac{m}{m + 1} \bullet \frac{1}{2m + 1}}$
$= \frac{2m^{2} + 2m + 1}{2m^{2} + 2m + 1} = 1$
$\Rightarrow \ \alpha + \beta = \frac{\pi}{4}$
(2) At $	heta = \frac{\pi}{4}$,
LHS = $3\ 	an{(45{^\circ} - 15{^\circ}) = 3	an{30{^\circ} = \sqrt{3}}}$
RHS = $	an{(45{^\circ} + 15{^\circ}) = 	an{60{^\circ} = \sqrt{3}}}$
$	herefore$ LHS = RHS
(3) Given $\sin^{2}{ax - \sin^{2}{(a - 1)x = \sin^{2}x}}$
$\Rightarrow \ \sin{(2a - 1)x\sin{(x) = \sin^{2}x}}$
$\Rightarrow \ \sin{x = 0\ and\ \sin{\ (2a - 1)x = \sin x}}$
$\Rightarrow \ x = n\pi\ and\ (2a - 1)x = n\pi + ( - 1)^{n}x$
Hence, option (a) is correct

Explanation

Reviewed by Dr. Abubacker

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