Question

State $T$ for true and $F$ for false.
(i) If $tan\,A=\frac{1-cos\,B}{sin\,B}$, then $tan2A=tanB$
(ii) The equality $sinA + sin2A + sin3A = 3$ holds for some real values of $A$.
(iii) $sin\, 10^{\circ}$ is greater than $cos \,10^{\circ}$.
(iv) $cos \frac{2\pi}{15} cos \frac{4\pi}{15} cos \frac{8\pi}{15} cos \frac{16\pi}{15}=\frac{1}{16}$

	(i)	(ii)	(iii)	(iv)
(a)	$F$	$F$	$T$	$T$
(b)	$F$	$T$	$F$	$T$
(c)	$T$	$T$	$F$	$F$
(d)	$T$	$F$	$F$	$T$

• A. (a)
• B. (b)
• C. (c)
• D. (d)

Answer 1

Answer: (D)

(i) True
We have, $tan\,A=\frac{1-cos\,B}{sin\,B}=\frac{1-1+2\,sin^{2} \frac{B}{2}}{2\,sin \frac{B}{2}\cdot cos \frac{B}{2}}=tan \frac{B}{2}\quad\ldots\left(i\right)$
Now, $tan\,2A=\frac{2\,tan\,A}{1-tan^{2}\,A}=\frac{2\cdot tan \frac{B}{2}}{1-tan^{2} \frac{B}{2}}=tan\,B$ (from $(i)$)

(ii) False
We have given that $sinA + sin2A + sin3A = 3$
It is possible only if $sinA$, $sin2A$, $sin3A$ each has a maximum value equal to one, which is not possible because angles are different.

(iii) False
$sin\,10^{\circ}=sin\left(90^{\circ}-80^{\circ}\right)$
$\Rightarrow sin\,10^{\circ}=cos80^{\circ}$
$\because cos80^{\circ} < cos10^{\circ}$. Hence, $sin10^{\circ} < cos10^{\circ}$
[$\because cosx$ is a decreasing function in $(0,\pi/2)$]

(iv) True
$L.H.S.=cos \frac{2\pi}{15} cos \frac{4\pi}{15} cos \frac{8\pi}{15} cos \frac{16\pi}{15}$

$=cos24^{\circ}\,cos48^{\circ}\,cos96^{\circ}\,cos192^{\circ}$

$=\frac{1}{16\,sin\,24^{\circ}}[\left(2\,sin\,24^{\circ}\,cos\,24^{\circ}\left(2\,cos\,48^{\circ}\right)\right)$

$(2cos96^{\circ}) (2cos192^{\circ})]$

$=\frac{1}{16\,sin\,24^{\circ}}\left[2\,sin\,48^{\circ }\,cos\,48^{\circ }\left(2\,cos\,96^{\circ }\right)\left(2cos192^{\circ}\right)\right]$

$=\frac{1}{16\,sin\,24^{\circ}} \left[\left(2\,sin\,96^{\circ}\,cos\,96^{\circ}\right)\left(2\,cos\,192^{\circ}\right)\right]$

$=\frac{1}{16\,sin\,24^{\circ}}\left[2\,sin\,192^{\circ}\,cos\,{}^{\circ}192^{\circ}\right]$

$=\frac{1}{16\,sin\,24^{\circ}} sin\,384^{\circ}=\frac{sin\left(360^{\circ}+24^{\circ}\right)}{16\,sin\,24^{\circ}}$

$=\frac{sin\,24^{\circ}}{16\,sin\,24^{\circ}}=\frac{1}{16}=R.H.S.$