Question

if 1 + sin square alpha equals to 3 sin alpha cos alpha then show that the value of tan Alpha is 1 or 1 by 2 ​

Answer 1

Given---> 1 + Sin²α = 3Sinα Cosα

To prove ---> Value of tanα is 1 or 1 / 2

Proof ---> ATQ ,

1 + Sin²α = 3 Sinα Cosα

Multiplying equation by 2 we get

=> 2 + ( 2Sin²α ) = 3 ( 2Sinα Cosα )

We have two formulee as follows

(1) Cos2A = 1 - 2Sin²A

2Sin²A = 1 - Cos2A

(2) Sin2A = 2 SinA CosA

Using these two formulee here we get

=> 2 + ( 1 - Cos2α ) = 3 Sin2α

=> 2 + 1 - Cos2α = 3 Sin2α

=> 3 - Cos2α = 3Sin2α

=> 3 Sin2α + Cos2α - 3 = 0

We have two formulee we get

Sin2A = 2 tanA / ( 1 + tan²A )

Cos2A = ( 1 - tan²A ) / (1 + tan²A )

Using these formulee here we get

=> 3 (2tanα /1 +tan²α) + (1-tan²α / 1 + tan²α ) -3 =0

Multiplying whole equation by ( 1 + tan²α ) we get

=> 3(2tanα ) + (1 - tan²α ) - 3 ( 1 + tan²α ) = 0

=> 6 tanα + 1 - tan²α - 3 - 3 tan²α = 0

=> - 4 tan²α + 6 tanα - 2 = 0

=> 2 tan²α - 3 tanα + 1 = 0

Using the method of splitting the middle term we get

=> 2 tan²α - ( 2 + 1 ) tanα + 1 = 0

=> 2 tan²α - 2 tanα - tanα + 1 = 0

Taking 2tanα common from first two term and (-1) common from next two terms we get

=> 2 tanα ( tanα - 1 ) - 1 ( tanα - 1 ) = 0

Taking ( tanα - 1 ) commn , we get

=> ( tanα - 1 ) ( 2tanα - 1 ) = 0

If tanα - 1 = 0

=> tanα = 1

If 2 tanα - 1 = 0

=> 2 tanα = 1

=> tanα = 1 / 2

Answer 2

Answer:

Here is your answer .

$\cot( \alpha ) = 2 \\ \tan( \alpha ) = \frac{1}{2}$

and

$\cot( \alpha ) = 1 \\ \tan( \alpha ) = 1$