Question

$\frac{\sqrt{1-co{s}^{2}A}}{\sqrt{se{c}^{2}A-1}}$$\times$$\frac{\sqrt{1+ta{n}^{2}A}}{\sqrt{1-si{n}^{2}A}}=$

• A. $tanA$
• B. $secA$
• C. $cosA$
• D. $cosecA$

Answer 1

The correct option is B $secA$

By Trigonometric identities
i)- $si{n}^{2}A+co{s}^{2}A=1$
$\Rightarrow si{n}^{2}A=1-co{s}^{2}A$

ii)- $se{c}^{2}A-ta{n}^{2}A=1$
(or) $se{c}^{2}A=1+ta{n}^{2}A$
(or) $ta{n}^{2}A=se{c}^{2}A-1$

$\therefore$ $\frac{\sqrt{1-co{s}^{2}A}}{\sqrt{se{c}^{2}A-1}}$ = $\frac{\sqrt{si{n}^{2}A}}{\sqrt{ta{n}^{2}A}}$ = $\frac{sinA}{tanA}$ ---- (1)

and $\frac{\sqrt{1+ta{n}^{2}A}}{\sqrt{1-si{n}^{2}A}}$ = $\frac{\sqrt{se{c}^{2}A}}{\sqrt{co{s}^{2}A}}$ = $\frac{secA}{cosA}$ ----(2)

From (1) and (2), we get,

$\frac{\sqrt{1-co{s}^{2}A}}{\sqrt{se{c}^{2}A-1}}$$\times$$\frac{\sqrt{1+ta{n}^{2}A}}{\sqrt{1-si{n}^{2}A}}$ = $\frac{sinA}{tanA}$ $\times$$\frac{secA}{cosA}$

= $\frac{sinA}{cosA}$ $\times$$\frac{secA}{tanA}$ [on rearranging]

= ($tanA$) $\times$$\frac{secA}{tanA}$ = $secA$

Hence,
$\frac{\sqrt{1-co{s}^{2}A}}{\sqrt{se{c}^{2}A-1}}\times \frac{\sqrt{1+ta{n}^{2}A}}{\sqrt{1-si{n}^{2}A}}=secA$