Question

if
$1 + \sin ^{2} x = 3 sinx \times cosx$
then prove that
$tanx = 1 \: or \: tanx = \frac{1}{2}$
​

Answer 1

Step-by-step explanation:

hope this helps you plz mark as brainliest...

Answer 2

||✪✪ QUESTION ✪✪||

If 1+ sin²x = 3sinx*cosx then prove that tanx = 1 or 1/2..?

|| ✰✰ ANSWER ✰✰ ||

⟿ 1 + sin²x = 3 * sinx * cosx

Dividing Both sides by cos²x, We get,

⟿ (1/cos²x) + (sin²x/cos²x) = 3 * (sinx/cos)

Now, putting (1/cos²x) = sec²x , and (sinx/cosx) = Tanx ,

⟿ sec²x + tan²x = 3 * tanx

Putting now sec²x = 1 + Tan²x , we get,

⟿ 1 + 2tan²x = 3tanx

⟿ 2tan²x - 3tanx + 1 = 0

Splitting The Middle Term now,

⟿ 2tan²x - 2tanx - tanx + 1 = 0

⟿ 2tanx(tanx -1) - 1(tanx - 1) = 0

⟿ (2tanx - 1)(tanx - 1) = 0

Putting both Equal to zero now,

☛ (2tanx - 1) = 0

☛ 2tanx = 1

☛ Tanx = (1/2)

OR,

☛ (Tanx -1) = 0

☛ Tanx = 1

☙☙ HENCE PROVED . ☙☙

∴ We can say that , If 1+ sin²x = 3sinx*cosx then tanx = 1 or (1/2)..