Question

prove that 1 - 2 cos square X = tan Square X - 1 / tan Square X + 1​

Answer 1

||✪✪ QUESTION ✪✪||

1 - 2cos²x = (tan²x - 1) / (tan²x + 1)

|| ✰✰ ANSWER ✰✰ ||

Taking LHS, we get,

→ (tan²x - 1) / (tan²x + 1)

Putting Tanx = sinx/cosx now,

→ (sin²x/cos²x) - 1/ (sin²x/cos²x) + 1 =

→ [ (sin²x - cos²x)/cos²x ] /[ ( sin²x + cos²x ) / cos²x ]

→ sin²x - cos²x /sin²x + cos²x

putting sin²x - cos²x = 1 in Denominator now,

→ sin²x - cos²x

Taking (-1) common now,

→ (-1)(cos²x - sin²x)

Putting cos²x - sin²x = cos2x now,

→ (-1) * cos2x

Putting cos2x = 2cos²x - 1 now,

→ (-1)(2cos²x - 1)

→ 1 - 2cos²x = ❁❁ RHS ❁❁

✪✪ Hence Proved ✪✪

Answer 2

$\huge\bold{Question}$

Prove that 1 - 2 cos² X = tan² X - 1 / tan² X + 1

$\huge\bold{Answer}$

By taking LHS, we get :-

$\implies \sf{ \frac{ (tan ^{2} x - 1) }{ (tan ^{2} x + 1)}}$

Putting substituting value of Tan x = sin x / cos x

= (sin²x/cos²x) - 1/ (sin²x/cos²x) + 1

= [ (sin²x - cos²x)/cos²x ] /[ ( sin²x + cos²x ) / cos²x ]

= sin²x - cos²x /sin²x + cos²x

•°• putting sin²x - cos²x = 1 in Denominator

________________( sin²x - cos²x = 1)

= sin²x - cos²x

Now taking - 1 common from the above eqn

= (-1) × (cos²x - sin²x)

Now by substituting the known value cos²x - sin²x = cos2x

= (-1) × cos2x

Now by substituting the known value cos2x = 2cos²x - 1

= (-1)(2cos²x - 1)

= 1 - 2cos²x