Question

From the following figure,
find:
(i) $y$
(ii) $\sin x^o$
(iii) $(\sec x^o-\tan x^o)(\sec x^o+\tan x^o)$

Answer 1

Given that,

(i) $y=?$

(ii) $\sin x^2=?$

(iii) $(\sec x^o-\tan x^o)(\sec x^o+\tan x^o)=?$

According to given figure,

By Pythagoras theorem,

$(Hypotenuse{)}^2=(perpendicular{)}^2+(Base{)}^2$

$2^2=1^2+y^2$

$4-1=y^2$

(i) $y=\sqrt{3}$

(ii) $\sin x^o=\frac{perpendicular}{Hypotenuse}$

$\sin x^o=\frac{y}{2}$

$\sin x^o=\frac{\sqrt{3}}{2}$

(iii) $(\sec x^o-\tan x^o)(\sec x^o+\tan x^o)$

$\Rightarrow {\sec }^2{x}^o-{\tan }^2{x}^o$

$\therefore (A-B)(A+B)=A^2=B^2$

So, By part (ii)

$\sin x^o=\frac{\sqrt{3}}{2}$

$\sin x^o=\sin {60}^o$

$x^o={60}^o$ put (iii)

${\sec }^2{x}^o-{\tan }^2{x}^o$

$\Rightarrow {\sec }^2{60}^o-{\tan }^2{60}^o$

$\Rightarrow (2{)}^2-(\sqrt{3}{)}^2$

$\Rightarrow 4-3$

$\Rightarrow 1$

${\sec }^2{x}^o-{\tan }^2{x}^o=1$

Hence, this is the answer.