Question

Prove each of the following identities:

1. (cosec2θ – 1) sin2θ = cos2θ

2. sin4A + sin2A cos2A = sin2A

3. cos2θ (1 + tan2θ) = 1

4. (1 + tan2θ) sin2θ = tan2θ.

Answer 1

Plzz post one question at one time.....mate

Answer 2

Answer:

The proof of all the identities are derived below

Step-by-step explanation:

The proof of all the identities is written as follows

1.

$LHS = (cosec {}^{2} θ - 1)sin {}^{2} θ \\ = ( \frac{1}{sin {}^{2} θ} - 1)sin {}^{2} θ \\ = 1 - sin {}^{2} θ = cos {}^{2} θ = RHS \\ = > LHS = RHS$

Hence,the first identity is proved.

2.

$LHS= sin {}^{4} A + sin {}^{2} Acos {}^{2} A \\ adding \: and \: subtracting \: some \: terms \: we \: get \\ = sin {}^{4} A + sin {}^{2} Acos {}^{2} A + sin {}^{2} Acos {}^{2} A + cos{}^{4} A - (sin {}^{2} Acos {}^{2} A + cos{}^{4} A ) \\ = ( {sin}^{2} A) {}^{2} + 2(sin {}^{2} A)(cos {}^{2} A) + (cos {}^{2} A) {}^{2} - (sin {}^{2} Acos {}^{2} A + cos{}^{4} A ) \\ = (sin {}^{2} A + cos {}^{2} A) {}^{2} - (sin {}^{2} Acos {}^{2} A + cos{}^{4} A )$

On simplifying we get

$LHS = (1 - cos {}^{4}A) - (sin {}^{2} Acos {}^{2} A) \\ = (1 - cos {}^{2}A)(1 + cos {}^{2} A) - (sin {}^{2} Acos {}^{2} A) \\ = sin {}^{2}A(1 + cos {}^{2} A) - (sin {}^{2} Acos {}^{2} A) \\ = sin {}^{2} A(1 + cos {}^{2} A - cos {}^{2} A) \\ = sin {}^{2} A = RHS \\ = > LHS = RHS$

Hence the identity is proved

3.

$LHS = cos {}^{2} θ(1 + tan {}^{2} θ) \\ = cos {}^{2} θ + cos {}^{2} θ \times \frac{sin {}^{2}θ }{cos {}^{2}θ } \\ = cos {}^{2} θ + sin {}^{2} θ = 1 = RHS \\ = > LHS = RHS$

Hence the identity is proved

4.

$LHS = (1 + tan {}^{2} θ)sin {}^{2}θ \\ = sin {}^{2} θ + \frac{sin {}^{4}θ }{cos {}^{2}θ } \\ = \frac{sin {}^{2}θcos {}^{2}θ + sin {}^{4} θ }{cos {}^{2}θ } \\ = \frac{sin {}^{2} θ(cos {}^{2} θ + sin {}^{2}θ) }{cos {}^{2}θ } \\ = \frac{sin {}^{2} θ}{cos {}^{2}θ } = tan {}^{2} θ = RHS \\ = > LHS = RHS$

Hence,the identity is proved.

Therefore,all the identities are proved

#SPJ3