Question

State $T$ for true and $F$ for false.
(i) If $tanA=\frac{1-cosB}{sinB}$, then $tan2A=tanB$
(ii) The equality $sinA+sin2A+sin3A=3$ holds for some real values of $A$.
(iii) $sin10^{\circ }$ is greater than $cos10^{\circ }$.
(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, $tanA=\frac{1-cosB}{sinB}=\frac{1-1+2si{n}^2\frac{B}{2}}{2sin\frac{B}{2}\cdot cos\frac{B}{2}}=tan\frac{B}{2}\dots \left(i\right)$
Now, $tan2A=\frac{2tanA}{1-ta{n}^{2}A}=\frac{2\cdot tan\frac{B}{2}}{1-ta{n}^2\frac{B}{2}}=tanB$ (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
$sin10^{\circ }=sin\left(90^{\circ }-80^{\circ }\right)$
$\Rightarrow sin10^{\circ }=cos80^{\circ }$
$\because cos80^{\circ }<cos10^{\circ }$. Hence, $sin10^{\circ }<cos10^{\circ }$
[$\because cosx$ is a decreasing function in $(0,\pi /2)$]

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

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

$=\frac{1}{16sin24^{\circ }}[\left(2sin24^{\circ }cos24^{\circ }\left(2cos48^{\circ }\right)\right)$

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

$=\frac{1}{16sin24^{\circ }}\left[2sin48^{\circ }cos48^{\circ }\left(2cos96^{\circ }\right)\left(2cos192^{\circ }\right)\right]$

$=\frac{1}{16sin24^{\circ }}\left[\left(2sin96^{\circ }cos96^{\circ }\right)\left(2cos192^{\circ }\right)\right]$

$=\frac{1}{16sin24^{\circ }}\left[2sin192^{\circ }cos{}^{\circ }192^{\circ }\right]$

$=\frac{1}{16sin24^{\circ }}sin384^{\circ }=\frac{sin\left(360^{\circ }+24^{\circ }\right)}{16sin24^{\circ }}$

$=\frac{sin24^{\circ }}{16sin24^{\circ }}=\frac{1}{16}=R.H.S.$