Question

If 1 + sin 0 = 3 sin 0 cos 0, prove that tan 0 = 1 or 1/2​

Answer 1

Step-by-step explanation:

$\sf 1 + {sin}^{2} \theta = 3sin \theta cos\theta$

$\sf Dividing\: by \:cos \theta\:$

$\sf {sec}^{2}\theta +{tan}^{2}\theta = 3tan \theta$

$\sf 1 + {tan}^{2}\theta + {tan}^{2}\theta - 3 tan \theta$

$\sf 2{tan}^{2}\theta - 3tan\theta + 1 = 0$

$\sf 2 {tan}^{2}\theta - 2tan\theta - tan\theta + 1 = 0$

$\sf ( 2tan\theta - 1 )( tan\theta - 1 ) = 0$

$\sf tan\theta = \frac {1}{2}$

$\sf tan\theta = 1$

Answer 2

Right  Question: 1+sin²a = 3sina×cosa

Given that

1+Sin²a= 3Sina×Cosa

Cos²a+Sin²a+Sin²a = 3Sina× Cosa [ as we know that  1 = Sin²+Cos A]

Cos²a+2Sin²a= 3Sina Cosa  ....(1)

DIVIDE (1) By COS²a we get

1+2tan²a = 3tan a

1+2tan²a - 3tan a = 0

(2tan a-1) ( tan a-1) = 0

2tan a -1 = 0. tan a -1= 0

2tan a = 1. tan a =1

therefore tan a = 1 or 1/2