Question

Write True' or False' and justify your answer in each of the following :
(i) The value of $\sin \mathrm{\theta }$ is $x+\frac{1}{x}$, where 'x' is a positive real number .
(ii) $\cos \mathrm{\theta }=\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{2\mathrm{ab}}$, where a and b are two distinct numbers such that ab > 0.

(iii) The value of ${\cos }^{2}23-{\sin }^{2}67$ is positive .

(iv) The value of the expression $\sin {80}^{°}-\cos {80}^{°}$ is negative .

(v) The value of $\sin \mathrm{\theta }+\cos \mathrm{\theta }$ is always greater than 1 .

Answer 1

(i)
$$\begin{gathered} \sin \theta =x+\frac{1}{x} \\ \Rightarrow -1\le x+\frac{1}{x}\le 1 \\ \Rightarrow x+\frac{1}{x}\le 1 \\ \Rightarrow x^2+1\le x \\ \Rightarrow x^2+1-x\le 0 \\ \mathrm{Take}x=1, \\ \Rightarrow 1+1-1\le 0 \\ \Rightarrow 1\le 0 \\ \mathrm{Which}\mathrm{is}\mathrm{false},\mathrm{so}x\mathrm{is}\mathrm{not}\mathrm{always}\mathrm{a}\mathrm{positive}\mathrm{real}\mathrm{number}. \\ \mathrm{The}\mathrm{given}\mathrm{statement}\mathrm{is}\mathrm{false}. \end{gathered}$$

(ii)
It is given that,
$$\begin{gathered} \cos \theta =\frac{a^2+b^2}{2ab} \\ \mathrm{Since},{\left(a-b\right)}^2=a^2+b^2-2ab>0\left[\because {\left(a-b\right)}^2>0,a\mathrm{and}b\mathrm{are}\mathrm{distint}\mathrm{numbers}\right] \\ \Rightarrow a^2+b^2>2ab \\ \Rightarrow \frac{a^2+b^2}{2ab}>1 \\ \Rightarrow \cos \theta >1 \\ \mathrm{But},-1\le \cos \theta \le 1 \\ \mathrm{Therefore},\mathrm{the}\mathrm{given}\mathrm{statement}\mathrm{is}\mathrm{false}. \end{gathered}$$

(iii)
$$\begin{gathered} {\cos }^{2}23°-{\sin }^{2}67° \\ ={\sin }^2\left(90°-23°\right)-{\sin }^{2}67° \\ ={\sin }^{2}67°-{\sin }^{2}67° \\ =0 \end{gathered}$$
Which is not positive, the given statement is false.

(iv)
Consider the table.

$\theta$	0º	30º	45º	60º	90º
sin$\theta$	0	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	1
cos$\theta$	1	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	0

Here,
$$\begin{gathered} \sin 60°-\cos 60°=\frac{\sqrt{3}}{2}-\frac{1}{2}>0 \\ \sin 90°-\cos 90°=1-0>0 \\ \mathrm{So},\sin 80°-\cos 80°>0\left(\sin \theta -\cos \theta \ge 0\forall 45°\le \theta \le 90°\right) \end{gathered}$$
Therefore, the given statement is false.

(v)
Consider the table.

$\theta$	0º	30º	45º	60º	90º
sin$\theta$	0	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	1
cos$\theta$	1	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	0

Here,
$$\begin{gathered} \sin 90°+\cos 90°=1+0=1,\mathrm{which}\mathrm{is}\mathrm{not}\mathrm{greater}\mathrm{than}1 \\ \mathrm{Therefore},\mathrm{the}\mathrm{given}\mathrm{statement}\mathrm{is}\mathrm{false}. \end{gathered}$$