Question

Ganit Prakash - Class X
Chapter : 23
5.
If cote theta=2, then let us determine the values of tan theta and sec theta and show that 1+tan^2 theta=sec^2 theta​

Answer 1

Step-by-step explanation:

Given, cot theta = 2.

=> cot theta = base / perpendicular = 2

=> tan theta = perpendicular / base = 1/2

$( {hypotenuse)}^{2} = ( {perpendicular)}^{2} + ( {base)}^{2}$

$( {h) }^{2} = ( {p)}^{2} + ( {b)}^{2}$

$= ( {1)}^{2} + ( {2)}^{2}$

$= 1 + 4$

$= 5$

$hypotenuse = \sqrt{5}$

$sec \: theta \: = \frac{hypotenuse}{base} = \frac{ \sqrt{5} }{2}$

therefore, L.H.S.,

$1 + {tan}^{2} theta = 1 + ( { \frac{1}{2} )}^{2}$

$= 1 + \frac{1}{4}$

$= \frac{4 + 1}{4}$

$= \frac{5}{4}$

R. H. S,

${sec}^{2} theta = ( { \frac{ \sqrt{5} }{2} )}^{2}$

$= \frac{5}{4}$

therefore, L. H. S = R. H. S (proved)